Question

answer the questions #3 - 6 given the functions below:
$g(x)=10 - x$, $f(x)=x^{3}$, $h(x)=-\frac{1}{2}x$, $j(x)=-(x + 4)^{2}+1$

1. find the average rate of change of $f(x)$ on the interval $-3,1$.

answer:________

1. the graph of $j(x)$ is shown below. on what interval(s) is $j(x)$ decreasing?

answer:________

1. find the value of:

$h(-5)*g(3)$
answer:________

1. if the domain of $j(x)$ is restricted to $3\leq x\leq7$, what is the range of $g(x)$ on this interval?

answer:________

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = f(x)\) on the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(a=-3\), \(b = 1\), and \(f(x)=x^{3}\).

Step2: Calculate \(f(a)\) and \(f(b)\)

When \(x=-3\), \(f(-3)=(-3)^{3}=-27\). When \(x = 1\), \(f(1)=1^{3}=1\).

Step3: Substitute into the formula

\(\frac{f(1)-f(-3)}{1-(-3)}=\frac{1-(-27)}{1 + 3}=\frac{1 + 27}{4}=\frac{28}{4}=7\).

Step4: Analyze the function \(j(x)=-(x + 4)^{2}+1\)

The function \(j(x)\) is a parabola in the form \(y=a(x - h)^{2}+k\) (here \(a=-1\), \(h=-4\), \(k = 1\)). Since \(a=-1<0\), the parabola opens down - ward. The vertex is at \((-4,1)\). A parabola \(y = a(x - h)^{2}+k\) is decreasing for \(x>h\) when \(a<0\). So \(j(x)\) is decreasing on the interval \((-4,\infty)\).

Step5: Calculate \(h(-5)\) and \(g(3)\)

Given \(h(x)=-\frac{1}{2}x\), then \(h(-5)=-\frac{1}{2}\times(-5)=\frac{5}{2}\). Given \(g(x)=10 - x\), then \(g(3)=10-3 = 7\). So \(h(-5)\times g(3)=\frac{5}{2}\times7=\frac{35}{2}=17.5\).

Step6: Analyze the function \(g(x)=10 - x\)

The function \(g(x)=10 - x\) is a linear function with slope \(m=-1\). Since \(m=-1<0\), the function is decreasing. When \(x = 3\), \(g(3)=10-3 = 7\). When \(x = 7\), \(g(7)=10 - 7=3\). So the range of \(g(x)\) for \(3\leq x\leq7\) is \([3,7]\).

Answer:

1. \(7\)
2. \((-4,\infty)\)
3. \(\frac{35}{2}\) (or \(17.5\))
4. \([3,7]\)