For in GOD we live, and move, and have our being.

- Acts 17:28

The Joy of a Teacher is the Success of his Students.

- Samuel Chukwuemeka

**
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 $

$ 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} $

$ (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

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

(b.) Calculate the

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

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.

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

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

$15 * (64 + 4) = 15 * 68 = \$1020$

Interesting application of Quadratic Functions!

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 $19 * 52 = \$988$

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 = $16 * 64 = \$1024$

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

$15 * (64 + 4) = 15 * 68 = \$1020$

Interesting application of Quadratic Functions!