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# Solved Examples on the Inverse of Functions

Determine the inverse of these functions.
Test for inverses.

(1.) $7x - 3$

$f(x) = 7x - 3 \\[3ex] y = 7x - 3 \\[3ex] Interchange\:\: x\:\: and\:\: y \\[3ex] x = 7y - 3 \\[3ex] Solve\:\: for\:\: y \\[3ex] 7y - 3 = x \\[3ex] 7y = x + 3 \\[3ex] y = \dfrac{x + 3}{7} \\[5ex] \therefore f^{-1}(x) = \dfrac{x + 3}{7}$

Test for Inverses
$(f \circ f^{-1})(x) \\[3ex] = f(f^{-1}(x)) \\[3ex] = f\left(\dfrac{x + 3}{7}\right) \\[5ex] But\:\: f(x) = 7x - 3 \\[5ex] So\:\: f\left(\dfrac{x + 3}{7}\right) = 7\left(\dfrac{x + 3}{7}\right) - 3 \\[5ex] = x + 3 - 3 \\[5ex] = x \\[7ex] (f^{-1} \circ f)(x) \\[3ex] = f^{-1}(f(x)) \\[3ex] = f^{-1}(7x - 3) \\[3ex] But\:\: f^{-1}(x) = \dfrac{x + 3}{7} \\[5ex] So\:\: f^{-1}(7x - 3) = \dfrac{(7x - 3) + 3}{7} \\[5ex] = \dfrac{7x - 3 + 3}{7} \\[5ex] = \dfrac{7x}{7} \\[5ex] = x \\[3ex] (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$
(2.) A ball is thrown directly upward from a height of $2\:\: ft$ with an initial velocity of $28\:\: ft/sec$.
The function: $h(t) = -16t^2 + 28t + 2$ gives the height of the ball in $feet$, $t$ seconds after it has been thrown.
(a.) Determine the time at which the ball reaches its maximum height.
(b.) Calculate the maximum height.

Notice the bold letter "maximum" in the question
The question is asking for "vertex"

$h(t) = -16t^2 + 28t + 2 \\[3ex] f(x) = ax^2 + bx + c \\[3ex] a = -16, b = 28, c = 2 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{28}{2(-16)} \\[5ex] -\dfrac{b}{2a} = \dfrac{-28}{-32} = \dfrac{7}{8} \\[5ex] f\left(-\dfrac{b}{2a}\right) = f\left(\dfrac{7}{8}\right) \\[5ex] But\:\: h(t) = -16t^2 + 28t + 2 \\[3ex] h\left(\dfrac{7}{8}\right) = -16\left(\dfrac{7}{8}\right)^2 + 28\left(\dfrac{7}{8}\right) + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -16\left(\dfrac{7}{8}\right)\left(\dfrac{7}{8}\right) + 7\left(\dfrac{7}{2}\right) + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -2\left(\dfrac{7}{1}\right)\left(\dfrac{7}{8}\right) + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -1\left(\dfrac{7}{1}\right)\left(\dfrac{7}{4}\right) + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -\dfrac{49}{4} + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -\dfrac{49}{4} + \dfrac{98}{4} + \dfrac{8}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{-49 + 98 + 8}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{57}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{57}{4} \\[5ex]$ The time to reach the maximum height is $\dfrac{7}{8} = 0.875$ seconds

The maximum height is $\dfrac{57}{4} = 14.25$ feet
(3.) JAMB, NGA A trader realizes $10x - x^2$ naira profit from the sale of $x$ bags of corn.
How many bags will give him the maximum profit?

We can do this question in two ways.
You can use any of the methods to solve it.

First Method: Vertex Formula - because the profit is a quadratic function
For this question:
the number of bags that will give the maximum profit = $x$ coordinate of the vertex
the maximum profit = $y$ coordinate of the vertex

$y = 10x - x^2 \\[3ex] y = -x^2 + 10x \\[3ex] y = ax^2 + bx + c \\[3ex] a = -1, b = 10 \\[3ex] x-coordinate\:\: of\:\: Vertex = -\dfrac{b}{2a} \\[5ex] = \dfrac{-10}{2(-1)} \\[5ex] = \dfrac{-10}{-2} \\[5ex] = 5 \\[3ex]$ The sale of $5$ bags will give the maximum profit

Second Method: Differential Calculus
One of the application of derivatives is in calculating maxima and minima of functions
You need to review Differential Calculus before using this method.

$y = 10x - x^2 \\[3ex] \dfrac{dy}{dx} = 10 - 2x \\[3ex] 10 - 2x = 0 \\[3ex] 10 = 2x \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[3ex] x = 5 \\[3ex]$ The sale of $5$ bags will give the maximum profit
(4.) ACT A sporting-goods store sells baseball caps for $\$22$each. At this price,$40$caps are sold per week. For every$\$1$ decrease in price, the store will sell $4$ more caps per week.
The store will adjust the price to maximize revenue.
What will be the maximum possible revenue for $1$ week?
(Note: The revenue equals the number of caps sold times the price per cap.)

Normal revenue for $1$ week = $40 * 22 = \$880$But, the store wants to make more money. For every$\$1$ decrease in price, the store will sell $4$ more caps per week.
For $22 - 1 = 21$, the store will sell $40 + 4 = 44$ caps
The company will sell $4$ more caps for a $\$1$decrease in price Revenue would be$21 * 44 = \$924$
For $22 - 1(2) = 20$, the store will sell $40 + 4(2) = 40 + 8 = 48$ caps
The company will sell $8$ more caps for a $\$2$decrease in price Revenue would be$20 * 48 = \$960$
For $22 - 1(3) = 19$, the store will sell $40 + 4(3) = 40 + 12 = 52$ caps
The company will sell $12$ more caps for a $\$3$decrease in price Revenue would be$19 * 52 = \$988$

Student: That would exactly take more than a minute to solve.
ACT questions typically should be solved at at an average of $1$ question per minute.
Teacher: I expected that comment.
I did those steps to explain the question to you.
I was doing it the Arithmetic method.
Let us now do it algebraically.

So;
For each $\$x$decrease in price, the company will sell$4x$more For$22 - 1(x) = 22 - x$, the store will sell$40 + 4(x) = 40 + 4x$caps Revenue would be$(22 - x) * (40 + 4x)$Let Revenue =$R R = (22 - x)(40 + 4x) \\[3ex] = 880 + 88x - 40x - 4x^2 \\[3ex] = 880 + 48x - 4x^2 \\[3ex] = -4x^2 + 48x + 880 \\[3ex] R = -4x^2 + 48x + 880 \\[3ex] y = ax^2 + bx + c \\[3ex] a = -4, b = 48 \\[3ex] x-coordinate\:\: of\:\: vertex = -\dfrac{b}{2a} \\[5ex] = -\dfrac{48}{2(-4)} = \dfrac{-48}{-8} = 6 \\[5ex] $The store has to decrease the price by$\$6$
This means that the store has to sell each cap for $22 - 6 = \$16$By doing so, the store will sell$40 + 4(6) = 40 + 24 = 64$caps Revenue =$16 * 64 = \$1024$

Student: So, this is the maximum revenue the store can get?
Teacher: That is correct.
Student: What if the store sells each cap for $\$15$? Teacher: Go ahead and find the revenue Student: Okay...it will be$15 * (64 + 4) = 15 * 68 = \$1020$