Question

1. let f be the function given by $f(x) = x^3 - 2x^2 + 5x - 16$. for what value of x in the closed interval 0,5 does the instantaneous rate of change of f equal the average rate of change of f over that interval?

a 0
b 5/3
c 5/2
d 3
e 5

Explanation:

Step1: Find average rate of change

The average rate of change on $[a,b]$ is $\frac{f(b)-f(a)}{b-a}$.
For $a=0$, $b=5$:
$f(0)=0^3-2(0)^2+5(0)-16=-16$
$f(5)=5^3-2(5)^2+5(5)-16=125-50+25-16=84$
$\text{Average rate}=\frac{84-(-16)}{5-0}=\frac{100}{5}=20$

Step2: Find derivative (instantaneous rate)

The derivative of $f(x)=x^3-2x^2+5x-16$ is:
$f'(x)=3x^2-4x+5$

Step3: Set rates equal, solve for x

Set $f'(x)=20$:
$3x^2-4x+5=20$
$3x^2-4x-15=0$
Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=3$, $b=-4$, $c=-15$:
$x=\frac{4\pm\sqrt{(-4)^2-4(3)(-15)}}{2(3)}=\frac{4\pm\sqrt{16+180}}{6}=\frac{4\pm\sqrt{196}}{6}=\frac{4\pm14}{6}$
Two solutions:
$x=\frac{4+14}{6}=\frac{18}{6}=3$, $x=\frac{4-14}{6}=\frac{-10}{6}=-\frac{5}{3}$
Only $x=3$ is in $[0,5]$.

Answer:

D. 3