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Functions Basics
What Is a Function?
A function is a rule that assigns exactly one output to every valid input. Think of it as a machine: you feed in a number, it applies a fixed process, and spits out a result.
Unique Output: Each input can only produce one output. If inputting 3 gives both 7 and 9, that rule is not a function.
Multiple Inputs, Same Output: Different inputs can share the same output. $f(2) = 5$ and $f(4) = 5$ is perfectly valid.
Process-Oriented: A function describes how to transform inputs, not just a static set of pairs.
Function notation uses $f(x)$ (read 'f of x') to name the rule and show where the input goes. The letter $f$ is the name of the rule, $x$ is the input, and $f(x)$ is the output value.
$$f(x) = \text{expression in } x$$
The notation packages the rule name, the input variable, and the output into one compact symbol.
$f$=Name of the function (can be any letter: $g$, $h$, $P$, $C$, etc.)(—)
$x$=The input variable (placeholder for the value you plug in)(—)
$f(x)$=The output value — the result after the rule is applied to $x$(depends on context)
$f(3)$
→Evaluate the rule with $x = 3$
$f(a + b)$
→Evaluate the rule with $x = a + b$ — substitute the entire expression
Naming Flexibility: Different letters signal meaning in context — $C(q)$ for cost as a function of quantity, $P(t)$ for profit as a function of time.
Not Multiplication: $f(x)$ does NOT mean $f \times x$. It means 'the value of function $f$ at input $x$'.
Example Function: $$g(x) = x^2 - 3x + 2$$
Domain and Range
The domain of a function is the set of all valid input values — the numbers you're allowed to plug in without breaking the rule.
Real-World Restrictions: If $f(q)$ represents the cost of producing $q$ items, the domain might be $q \geq 0$ (you can't produce negative items) or $q \in \{0, 1, 2, 3, \ldots\}$ (whole items only).
Mathematical Restrictions: Denominators cannot be zero, and square roots (of real numbers) cannot have negative radicands.
Stating the Domain: Write as an inequality ($x \geq 2$), an interval ($[2, \infty)$), or in set notation ($\{x \in \mathbb{R} \mid x \geq 2\}$).
The range of a function is the set of all possible output values — every result the function can actually produce when you feed it valid inputs from the domain.
Dependent on Domain: The range depends on which inputs you allow. Restrict the domain and the range changes too.
Finding the Range: Plug in domain values and collect all outputs. For continuous functions, look for minimum/maximum values.
Not Every Real Number: Even if the codomain is 'all real numbers', the range might be smaller — e.g., $f(x) = x^2$ never produces negative outputs.
Domain Restriction Quick Reference
•
Denominator: expression under fraction ≠ 0
•
Square root: expression under root ≥ 0
•
Logarithm: expression inside log > 0
•
Real-world context: non-negative values, whole numbers, or specific intervals
Evaluating Functions
Evaluating a function means substituting a specific number (or expression) for the input variable and simplifying to get the output.
$$f(a) = \text{replace every } x \text{ with } a, \text{ then simplify}$$
Evaluation is mechanical: swap out the variable, then compute.
$f(a)$=The function evaluated at the specific input $a$(depends on context)
$a$=Any number or expression substituted for $x$(—)
$f(-x)$
→Substitute $-x$ for every $x$ — useful for checking even/odd symmetry
$f(x + h)$
→Substitute $x + h$ for every $x$ — needed for difference quotients
Substitute Everything: Every occurrence of $x$ gets replaced — including inside exponents, fractions, and parentheses.
Simplify Fully: After substitution, carry out all arithmetic. $f(3) = (3)^2 - 2(3) + 1 = 9 - 6 + 1 = 4$.
Expression Inputs: You can evaluate at expressions, not just numbers. $f(a+b)$ means substitute $(a+b)$ everywhere $x$ appears.
Linear Functions
A linear function produces a constant rate of change — the output changes by the same amount for every unit increase in input. Its graph is always a straight line.
$$f(x) = mx + b$$
The output equals the slope times the input plus a starting value (y-intercept).
$m$=Slope — the rate of change (rise over run)(units of output per unit of input)
$b$=Y-intercept — the output when $x = 0$(same as output)
$x$=The input variable(depends on context)
$f(x)$=The output value(depends on context)
$m = 0$
→Constant function — $f(x) = b$, a horizontal line
$b = 0$
→Direct proportionality — $f(x) = mx$, the line passes through the origin
$m > 0$
→Increasing function — output rises as input grows
$m < 0$
→Decreasing function — output falls as input grows
Slope as Rate: If $f(t) = 40t + 100$ models savings over months, you save $40/month (slope) starting from $100 (intercept).
Finding Slope from Two Points: Given $(x_1, y_1)$ and $(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Domain and Range: For a linear function, both are typically all real numbers unless real-world constraints apply.
Quadratic Functions
A quadratic function has a squared term, giving it a parabolic (U-shaped or inverted U-shaped) graph. The output changes at an accelerating (or decelerating) rate.
$$f(x) = ax^2 + bx + c$$
A polynomial of degree 2 that produces a parabola when graphed.
$a$=Leading coefficient — controls whether the parabola opens up or down and how wide/narrow it is(—)
$b$=Linear coefficient — affects the horizontal position of the vertex(—)
$c$=Constant term — the y-intercept ($f(0) = c$)(same as output)
$a > 0$
→Parabola opens upward — vertex is the minimum point
$a < 0$
→Parabola opens downward — vertex is the maximum point
$a = 0$
→Degenerates to a linear function $f(x) = bx + c$
Direction: The sign of $a$ tells you which way the parabola opens — positive means a smile (minimum), negative means a frown (maximum).
Width: Larger $|a|$ makes a narrower parabola (steeper curve); smaller $|a|$ makes a wider, flatter one.
Y-Intercept: Always at $(0, c)$ — plug in $x = 0$ to find where the graph crosses the y-axis.
The vertex of a parabola is its highest or lowest point. Its x-coordinate is found using a simple formula derived from the coefficients.
$$x_v = -\frac{b}{2a}, \quad f(x_v) = c - \frac{b^2}{4a}$$
The vertex sits at the axis of symmetry — a vertical line $x = -\frac{b}{2a}$ that divides the parabola into two mirror halves.
$x_v$=X-coordinate of the vertex (input value at the turning point)(same as input)
$f(x_v)$=Y-coordinate of the vertex (maximum or minimum output)(same as output)
$b = 0$
→Vertex is on the y-axis at $x = 0$, so the vertex is $(0, c)$
Maximum vs Minimum: If $a > 0$, the vertex is the minimum. If $a < 0$, the vertex is the maximum.
Axis of Symmetry: The vertical line $x = x_v$ is the mirror line — points equally distant from it have the same y-value.
Vertex Form: Rewriting as $f(x) = a(x - h)^2 + k$ directly reveals the vertex at $(h, k)$.
Piecewise Functions
A piecewise function uses different rules for different parts of its domain. Each rule applies only to inputs that satisfy a specific condition.
$$f(x) = \begin{cases} \text{rule}_1 & \text{if condition}_1 \\ \text{rule}_2 & \text{if condition}_2 \\ \vdots & \vdots \end{cases}$$
Different formulas govern different intervals or conditions of the input.
$\text{rule}_i$=The expression used for inputs satisfying condition $i$(—)
$\text{condition}_i$=The domain restriction for which rule $i$ applies(—)
Check Conditions First: To evaluate $f(a)$, check which condition $a$ satisfies, then use the matching rule.
Non-Overlapping Conditions: The conditions should cover the entire domain without overlapping — each input belongs to exactly one rule.
Real-World Use: Tax brackets, shipping costs, and bulk pricing are all piecewise — the rule changes at thresholds.
Composition of Functions
Function composition chains two functions together: the output of one becomes the input of the next. It represents applying one process after another.
$$(f \circ g)(x) = f(g(x))$$
First apply $g$ to $x$, then feed that result into $f$.
$(f \circ g)(x)$='f composed with g' — apply $g$ first, then $f$(depends on context)
$g(x)$=The inner function — evaluated first(—)
$f(\cdot)$=The outer function — applied to the result of $g(x)$(—)
$f \circ g \neq g \circ f$
→Composition is generally NOT commutative — order matters
Right to Left: $f(g(x))$ means work from the inside out — compute $g(x)$ first, then plug that into $f$.
Order Matters: If $f(x) = 2x$ and $g(x) = x + 3$, then $f(g(x)) = 2(x+3) = 2x + 6$ but $g(f(x)) = 2x + 3$ — different results.
Domain of Composition: The domain of $f \circ g$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
Application Strategy
Worked Examples
Problem Statement
A company's daily revenue in thousands of rupees is modeled by $R(q) = 8q - q^2$, where $q$ is the number of units produced (in hundreds). Find the revenue when 300 units are produced.
evaluating a function
Strategic Toolkit
$$R(q) = 8q - q^2$$
Step 1
Convert units
The input $q$ is in hundreds of units, so 300 units equals $q = 3$.
$q = \frac{300}{100} = 3$
Step 2
Substitute into the function
Replace every $q$ with 3 in the revenue formula.
$R(3) = 8(3) - (3)^2$
Step 3
Simplify
Compute the multiplication and subtraction to get the revenue.
$R(3) = 24 - 9 = 15$
Step 4
Interpret the result
The revenue is 15 thousand rupees when 300 units are produced.
$R(3) = 15 \text{ thousand rupees}$
Problem Statement
Find the domain of $f(x) = \frac{3x + 1}{x - 4}$.
domainevaluating a function
Strategic Toolkit
$$f(x) = \frac{3x + 1}{x - 4}$$
Step 1
Identify the restriction
The denominator cannot be zero. Set it equal to zero to find the forbidden value.
$x - 4 \neq 0$
Step 2
Solve for the excluded value
The value $x = 4$ makes the denominator zero, so it is excluded from the domain.
$x \neq 4$
Step 3
State the domain
All real numbers except 4 are valid inputs.
$\text{Domain} = \{x \in \mathbb{R} \mid x \neq 4\}$
Problem Statement
A parking lot charges a flat fee of Rs. 50 plus Rs. 30 per hour. Write the cost function and find the cost for 5 hours of parking.
linear function
Strategic Toolkit
$$C(h) = 30h + 50$$
Step 1
Identify slope and intercept
The per-hour charge (Rs. 30) is the slope, and the flat fee (Rs. 50) is the y-intercept.
$m = 30, \quad b = 50$
Step 2
Write the function
Substitute $m$ and $b$ into the linear function form $C(h) = mh + b$.
$C(h) = 30h + 50$
Step 3
Evaluate at 5 hours
Substitute $h = 5$ and compute the total cost.
$C(5) = 30(5) + 50 = 150 + 50 = 200$
Step 4
State the answer
Five hours of parking costs Rs. 200.
$C(5) = \text{Rs. } 200$
Problem Statement
A shopkeeper's daily profit (in rupees) for selling $x$ items is $P(x) = -2x^2 + 80x - 200$. Find the number of items that maximizes profit and the maximum profit.
quadratic functionvertex
Strategic Toolkit
$$x_v = -\frac{b}{2a}$$
Step 1
Identify coefficients
Extract $a$, $b$, and $c$ from the quadratic. Since $a < 0$, the parabola opens downward and the vertex gives the maximum.
$a = -2, \quad b = 80, \quad c = -200$
Step 2
Find the vertex x-coordinate
Use the vertex formula to find the input value that maximizes profit.
$x_v = -\frac{80}{2(-2)} = -\frac{80}{-4} = 20$
Step 3
Evaluate profit at the vertex
Substitute $x = 20$ into the profit function to find the maximum profit.
$P(20) = -2(20)^2 + 80(20) - 200 = -2(400) + 1600 - 200 = -800 + 1600 - 200$
Step 4
Simplify to get maximum profit
Complete the arithmetic to find the peak profit value.
$P(20) = 600$
Step 5
Interpret
Selling 20 items yields the maximum daily profit of Rs. 600.
$\text{Max profit} = \text{Rs. } 600 \text{ at } x = 20 \text{ items}$
Problem Statement
A delivery service charges: Rs. 100 for packages up to 2 kg, Rs. 150 for packages over 2 kg but up to 5 kg, and Rs. 250 for packages over 5 kg. The cost function is $C(w) = \begin{cases} 100 & 0 < w \leq 2 \\ 150 & 2 < w \leq 5 \\ 250 & w > 5 \end{cases}$. Find $C(1.5)$, $C(3)$, and $C(7)$.
piecewise functionevaluating a function
Strategic Toolkit
$$C(w) = \begin{cases} 100 & 0 < w \leq 2 \\ 150 & 2 < w \leq 5 \\ 250 & w > 5 \end{cases}$$
Step 1
Evaluate $C(1.5)$
Since $1.5$ satisfies $0 < w \leq 2$, use the first rule.
$C(1.5) = 100$
Step 2
Evaluate $C(3)$
Since $3$ satisfies $2 < w \leq 5$, use the second rule.
$C(3) = 150$
Step 3
Evaluate $C(7)$
Since $7 > 5$, use the third rule.
$C(7) = 250$
Problem Statement
Given $f(x) = 3x + 1$ and $g(x) = x^2$, find $(f \circ g)(4)$ and $(g \circ f)(4)$.
function compositionevaluating a function
Strategic Toolkit
$$(f \circ g)(x) = f(g(x))$$
Step 1
Find $(f \circ g)(4)$ — inner function first
Compute $g(4)$ first since $g$ is the inner function.
$g(4) = (4)^2 = 16$
Step 2
Apply outer function $f$
Feed the result $16$ into $f$.
$f(16) = 3(16) + 1 = 48 + 1 = 49$
Step 3
Find $(g \circ f)(4)$ — reverse order
This time, $f$ is the inner function. Compute $f(4)$ first.
$f(4) = 3(4) + 1 = 13$
Step 4
Apply outer function $g$
Feed $13$ into $g$.
$g(13) = (13)^2 = 169$
Step 5
Compare results
The two compositions give different results, confirming that order matters.
$(f \circ g)(4) = 49 \neq 169 = (g \circ f)(4)$
Problem Statement
Find the domain of $h(x) = \frac{\sqrt{x + 2}}{x - 1}$.
domain
Strategic Toolkit
$$h(x) = \frac{\sqrt{x + 2}}{x - 1}$$
Step 1
Restriction from the square root
The radicand must be non-negative for a real-valued output.
$x + 2 \geq 0 \implies x \geq -2$
Step 2
Restriction from the denominator
The denominator cannot be zero.
$x - 1 \neq 0 \implies x \neq 1$
Step 3
Combine restrictions
Both conditions must hold simultaneously: $x \geq -2$ and $x \neq 1$.
$\text{Domain} = [-2, 1) \cup (1, \infty)$
Problem Statement
Find the range of $f(x) = 2x^2 - 8x + 5$.
quadratic functionvertexrange
Strategic Toolkit
$$x_v = -\frac{b}{2a}$$
Step 1
Identify coefficients
Since $a = 2 > 0$, the parabola opens upward, so the vertex gives the minimum value.
$a = 2, \quad b = -8, \quad c = 5$
Step 2
Find the vertex x-coordinate
Apply the vertex formula.
$x_v = -\frac{-8}{2(2)} = \frac{8}{4} = 2$
Step 3
Evaluate $f$ at the vertex
Compute the minimum output value.
$f(2) = 2(2)^2 - 8(2) + 5 = 2(4) - 16 + 5 = 8 - 16 + 5 = -3$
Step 4
State the range
Since the parabola opens upward, the minimum is $-3$ and the function grows without bound above it.
$\text{Range} = [-3, \infty)$
Memorize formulas
Linear Function
Use when the rate of change is constant — straight-line relationships.
Formula
$f(x) = mx + b$
Quadratic Function (Standard Form)
Use for parabolic relationships — profit max, area problems, projectile motion.
Formula
$f(x) = ax^2 + bx + c$
Vertex x-Coordinate
Use to find the turning point (max or min) of a quadratic.
Formula
$x_v = -\frac{b}{2a}$
Slope from Two Points
Use to find the rate of change between any two points on a line.
Formula
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Function Composition
Use when one process feeds into another — apply inner function first.
Formula
$(f \circ g)(x) = f(g(x))$
Review
1
Which of the following sets of ordered pairs represents a function?
Ans
{(1, 2), (2, 3), (3, 4)}
Review
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Check each set for duplicate inputs
For each option, verify whether any x-value appears more than once with a different y-value.
Only {(1, 2), (2, 3), (3, 4)} has all unique inputs
Apply the function definition
A relation is a function if and only if every element of the domain is paired with exactly one element of the range.
$$\text{Only set 2 satisfies this condition}$$
2
What is the domain of $f(x) = \frac{3}{x - 5}$?
Ans
All real numbers except $x = 5$
Review
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Identify the restriction
For a rational function, set the denominator not equal to zero.
$$x - 5 \neq 0$$
Solve for $x$
The only excluded value is $x = 5$; everything else is valid.
$$\text{Domain} = \mathbb{R} \setminus \{5\}$$
3
If $f(x) = 4x - 7$, what is $f(3)$?
Ans
5
Review
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Substitute $x = 3$
Replace every $x$ in $f(x) = 4x - 7$ with the value 3.
$$f(3) = 4(3) - 7$$
Compute the result
Multiply first, then subtract.
$$f(3) = 12 - 7 = 5$$
4
A machine's output rate in items per hour depends on temperature $T$ (in °C) as $f(T) = \frac{120}{T - 25}$. At what temperature does this model break down?
Ans
25°C
Review
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Find where the denominator is zero
The model breaks down when $T - 25 = 0$.
$$T - 25 = 0 \implies T = 25$$
Interpret the result
At 25°C the formula produces division by zero, so the model is undefined there.
$$\text{The model is invalid at } T = 25°C$$
5
A taxi company charges a base fare of Rs. 200 plus Rs. 15 per kilometer. Which function represents the total cost $C$ (in rupees) for traveling $x$ kilometers?
Ans
$C(x) = 15x + 200$
Review
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Identify slope and intercept
The per-kilometer rate (Rs. 15) is the slope $m$, and the base fare (Rs. 200) is the y-intercept $b$.
$$m = 15, \quad b = 200$$
Write the linear function
Substitute into $f(x) = mx + b$.
$$C(x) = 15x + 200$$
6
What is the slope of the line passing through the points $(2, 5)$ and $(6, 13)$?
Ans
2
Review
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Apply the slope formula
Substitute the two points into $m = \frac{y_2 - y_1}{x_2 - x_1}$.
$$m = \frac{13 - 5}{6 - 2} = \frac{8}{4}$$
Simplify
Reduce the fraction to get the final slope.
$$m = 2$$
7
A ball is thrown upward and its height in meters after $t$ seconds is $h(t) = -5t^2 + 20t + 1$. What is the maximum height reached by the ball?
Ans
21 meters
Review
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Find the vertex $t$-coordinate
Use $t_v = -\frac{b}{2a}$ where $a = -5$ and $b = 20$.
$$t_v = \frac{-20}{2(-5)} = 2 \text{ seconds}$$
Evaluate $h(t)$ at the vertex
Substitute $t = 2$ into the height function to get the maximum height.
$$h(2) = -5(4) + 20(2) + 1 = 21 \text{ meters}$$
8
A salesperson earns a monthly commission of $C(s) = 0.08s + 2000$, where $s$ is total sales in rupees. If the salesperson earned Rs. 6,800 in commission last month, what were the total sales?
Ans
Rs. 60,000
Review
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Set up the equation
Equate the commission function to the given commission amount.
$$0.08s + 2000 = 6800$$
Solve for $s$
Isolate $s$ by subtracting 2000 and then dividing by 0.08.
$$0.08s = 4800 \implies s = \frac{4800}{0.08} = 60{,}000$$
9
A function is defined as $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \geq 0 \end{cases}$. What is $f(-3)$?
Ans
9
Review
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Determine which piece applies
Check which condition the input $x = -3$ satisfies.
$$-3 < 0 \implies \text{use } f(x) = x^2$$
Evaluate using the correct piece
Substitute $x = -3$ into $x^2$.
$$f(-3) = (-3)^2 = 9$$
10
A bakery's daily flour consumption in kg is $F(d) = 3d + 1$, where $d$ is dozens of loaves baked. The number of dozens baked depends on the day of the month $n$ as $d(n) = 5n$. How much flour is used on the 4th day of the month?
Ans
61 kg
Review
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Find dozens baked on day 4
Evaluate $d(n) = 5n$ at $n = 4$.
$$d(4) = 5(4) = 20 \text{ dozens}$$
Find flour consumption for 20 dozens
Evaluate $F(d) = 3d + 1$ at $d = 20$.
$$F(20) = 3(20) + 1 = 61 \text{ kg}$$
11
What is the range of $f(x) = x^2 - 4x + 3$?
Ans
$[-1, \infty)$
Review
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Complete the square
Rewrite $x^2 - 4x + 3$ in vertex form by completing the square on the $x$ terms.
$$f(x) = (x - 2)^2 - 1$$
Determine the range from vertex form
The squared term $(x-2)^2$ has a minimum of 0, so $f(x)$ has a minimum of $-1$ and extends upward without bound.
$$\text{Range} = [-1, \infty)$$
12
What is the domain of $f(x) = \frac{1}{\sqrt{x - 2}}$?
Ans
$x > 2$
Review
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Apply the square root restriction
The radicand must be non-negative: $x - 2 \geq 0$.
$$x \geq 2$$
Apply the denominator restriction
The square root in the denominator cannot be zero, so $\sqrt{x - 2} \neq 0$, which means $x - 2 \neq 0$.
$$x - 2 > 0 \implies x > 2$$
13
A machine bought for Rs. 80,000 depreciates by Rs. 5,000 each year. After how many years will its value be Rs. 35,000?
Ans
9 years
Review
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Set up the linear depreciation model
Starting value Rs. 80,000 decreases by Rs. 5,000 per year.
$$V(t) = 80000 - 5000t$$
Solve for the time when value equals Rs. 35,000
Set $V(t) = 35000$ and solve for $t$.
$$80000 - 5000t = 35000 \implies t = 9 \text{ years}$$
14
The vertex of a parabola is at $(3, -2)$ and it passes through the point $(1, 6)$. What is the equation of the parabola in standard form?
Ans
$y = 2x^2 - 12x + 16$
Review
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Write vertex form with unknown $a$
Use the vertex $(h, k) = (3, -2)$ in $y = a(x - h)^2 + k$.
$$y = a(x - 3)^2 - 2$$
Solve for $a$ using the given point
Substitute $(1, 6)$ into the vertex form equation.
$$6 = a(1-3)^2 - 2 \implies 8 = 4a \implies a = 2$$
Expand to standard form
Distribute and simplify to get $y = ax^2 + bx + c$.
$$y = 2x^2 - 12x + 16$$
15
If $(f \circ g)(x) = 4x^2 - 12x + 9$ and $g(x) = 2x - 3$, what is $f(x)$?
Ans
$f(x) = x^2$
Review
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Recognize the composition pattern
We know $f(g(x)) = 4x^2 - 12x + 9$ and $g(x) = 2x - 3$. Try to express the composition in terms of $g(x)$.
$$4x^2 - 12x + 9 = (2x - 3)^2 = [g(x)]^2$$
Identify $f$
Since $f(g(x)) = [g(x)]^2$, the outer function squares its input.
$$f(x) = x^2$$
16
A parking lot charges Rs. 50 for the first hour and Rs. 30 for each additional hour or part thereof. If Sara parks for 4.5 hours, what is her total charge?
Ans
Rs. 170
Review
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Identify the two pieces of the cost function
First hour is a flat Rs. 50. Each subsequent hour (or fraction) costs Rs. 30.
$$C(t) = \begin{cases} 50 & 0 < t \leq 1 \\ 50 + 30 \cdot \lceil t - 1 \rceil & t > 1 \end{cases}$$
Compute additional hours charged
For 4.5 hours, the additional time is 3.5 hours, which rounds up to 4 chargeable units.
$$\lceil 4.5 - 1 \rceil = \lceil 3.5 \rceil = 4$$
Calculate total charge
Add the first-hour charge to the additional-hour charges.
$$50 + 4 \times 30 = 170$$
17
A store's daily revenue in hundreds of rupees is $R(d) = 5d + 20$, where $d$ is the number of customers in dozens. The number of customers depends on the day's high temperature $t$ (in °C) as $d(t) = 2t + 1$. What is the revenue when the temperature is 20°C?
Ans
Rs. 22,500
Review
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Find the number of customers at 20°C
Evaluate $d(t) = 2t + 1$ at $t = 20$.
$$d(20) = 2(20) + 1 = 41 \text{ dozens}$$
Compute revenue from 41 dozen customers
Evaluate $R(d) = 5d + 20$ at $d = 41$.
$$R(41) = 5(41) + 20 = 225 \text{ (hundreds of rupees)}$$
Convert to actual rupees
Multiply by 100 to convert from hundreds of rupees to rupees.
$$\text{Revenue} = 225 \times 100 = \text{Rs. } 22{,}500$$
18
A company's profit in thousands of rupees is modeled by $P(x) = 2.5x - 15$, where $x$ is units sold in hundreds. What does $P(0)$ represent and what is its value?
Ans
A loss of Rs. 15,000 when no units are sold
Review
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Evaluate $P(0)$
Substitute $x = 0$ into the profit function.
$$P(0) = 2.5(0) - 15 = -15$$
Interpret the result
The value $-15$ in thousands of rupees means a loss of Rs. 15,000, occurring when no units are sold.
$$\text{At } x = 0 \text{ (zero sales), the company loses Rs. 15,000}$$
19
For the function $f(x) = -2x^2 + 8x - 5$, which statement is correct?
Ans
The maximum value of $f(x)$ is 3
Review
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Find the vertex
Use $x_v = -\frac{b}{2a}$ with $a = -2$, $b = 8$.
$$x_v = \frac{-8}{2(-2)} = 2$$
Evaluate $f$ at the vertex
Compute $f(2)$ to find the extreme value.
$$f(2) = -2(4) + 8(2) - 5 = 3$$
Determine whether it is a maximum or minimum
Since $a = -2 < 0$, the parabola opens downward, so the vertex is a maximum.
$$\text{Maximum value} = 3$$
20
If $f(x) = \frac{1}{x}$ and $g(x) = x - 3$, what is the domain of $(f \circ g)(x)$?
Ans
All real numbers except $x = 3$
Review
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Form the composition
Compute $(f \circ g)(x) = f(g(x))$ by substituting $g(x) = x - 3$ into $f$.
$$(f \circ g)(x) = \frac{1}{x - 3}$$
Find where the composition is undefined
Set the denominator equal to zero and exclude that value from the domain.
$$x - 3 \neq 0 \implies \text{Domain} = \mathbb{R} \setminus \{3\}$$
Question 1 of 20
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1
Which of the following sets of ordered pairs represents a function?