Question

quiz 4
question 1
classify this function
question 2
you are given four functions: $a(x)=x^{3}-5x$, $b(x)= - 6x - 1$, $c(x)=x^{6}+3$, and $d(x)=x^{2}+x$
which one is an even function?
question 3
what is the domain of the following function?
question 4
what is the inverse of the following function?
$f(x)=\frac{4x - 3}{5}$
question 5
find the zeroes of the parabola modelled by
$f(x)=4x^{2}+16x - 65$
question 6
find the optimal value of the parabola modelled by
$f(x)=4x^{2}+16x - 65$
question 7
find the optimal value of the parabola modelled by
$f(x)=-3(x - 7)(x + 23)$
question 8
find the zeroes of the parabola modelled by
$f(x)=2(x - 8)^{2}-50$

Explanation:

Step1: Identify function type for Question 1

The given function in Question 1 has discrete - valued points, so it is a discrete function.

Step2: Check for even - function in Question 2

For a function \(y = f(x)\) to be even, \(f(-x)=f(x)\).
For \(a(x)=x^{3}-5x\), \(a(-x)=(-x)^{3}-5(-x)=-x^{3} + 5x=-a(x)\), so it is odd.
For \(b(x)=-6x - 1\), \(b(-x)=6x - 1
eq b(x)\) and \(b(-x)
eq - b(x)\).
For \(c(x)=x^{6}+3\), \(c(-x)=(-x)^{6}+3=x^{6}+3 = c(x)\), so \(c(x)\) is even.
For \(d(x)=x^{2}+x\), \(d(-x)=x^{2}-x
eq d(x)\) and \(d(-x)
eq - d(x)\).

Step3: Determine domain for Question 3

The graph in Question 3 is a continuous function that extends infinitely in the \(x\) - direction. The domain is all real numbers, \((-\infty,\infty)\).

Step4: Find inverse for Question 4

Let \(y = f(x)=\frac{4x - 3}{5}\).
First, swap \(x\) and \(y\): \(x=\frac{4y - 3}{5}\).
Then solve for \(y\):
\[

$$\begin{align*} 5x&=4y-3\\ 4y&=5x + 3\\ y&=\frac{5x+3}{4} \end{align*}$$

\]
So \(f^{-1}(x)=\frac{5x + 3}{4}\).

Step5: Find zeroes for Question 5

For \(f(x)=4x^{2}+16x - 65\), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 4\), \(b = 16\), and \(c=-65\).
\[

$$\begin{align*} x&=\frac{-16\pm\sqrt{16^{2}-4\times4\times(-65)}}{2\times4}\\ &=\frac{-16\pm\sqrt{256 + 1040}}{8}\\ &=\frac{-16\pm\sqrt{1296}}{8}\\ &=\frac{-16\pm36}{8} \end{align*}$$

\]
\(x_1=\frac{-16 + 36}{8}=\frac{20}{8}=\frac{5}{2}\) and \(x_2=\frac{-16-36}{8}=\frac{-52}{8}=-\frac{13}{2}\)

Step6: Find optimal value for Question 6

For the parabola \(y = 4x^{2}+16x - 65\), the \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}=-\frac{16}{2\times4}=- 2\).
Substitute \(x = - 2\) into the function: \(y=4(-2)^{2}+16(-2)-65=16-32 - 65=-81\). The optimal value (minimum since \(a = 4>0\)) is \(-81\).

Step7: Find optimal value for Question 7

First, expand \(f(x)=-3(x - 7)(x + 23)=-3(x^{2}+23x-7x - 161)=-3(x^{2}+16x - 161)\)
The \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}=-\frac{16}{2\times1}=-8\)
Substitute \(x=-8\) into \(f(x)\): \(f(-8)=-3((-8)^{2}+16(-8)-161)=-3(64-128 - 161)=-3(-225)=675\)

Step8: Find zeroes for Question 8

For \(f(x)=2(x - 8)^{2}-50\), set \(f(x)=0\).
\[

$$\begin{align*} 2(x - 8)^{2}-50&=0\\ 2(x - 8)^{2}&=50\\ (x - 8)^{2}&=25\\ x-8&=\pm5 \end{align*}$$

\]
\(x_1=8 + 5=13\) and \(x_2=8-5 = 3\)

Answer:

Question 1: Discrete function
Question 2: \(c(x)=x^{6}+3\)
Question 3: \((-\infty,\infty)\)
Question 4: \(f^{-1}(x)=\frac{5x + 3}{4}\)
Question 5: \(x=\frac{5}{2},x=-\frac{13}{2}\)
Question 6: \(-81\)
Question 7: \(675\)
Question 8: \(x = 3,x = 13\)