CPM Homework Help : CALC3RD Problem 7-51

The point from problem 7-19 travels along the $x$ -axis so that at time $t$ its position is given by $s(t) = t^3 - 5t^2 + 4t$ , where $0 ≤ t ≤ 5$ . Calculate the average velocity of the point. At what time(s) during the interval was the point traveling at this average velocity?

Hint:

This is an application of the Mean Value Theorem of $f(x)$ (given $F(x)$ ).
In other words, where on the interval $[0,5]$ is the slope of the tangent the same as the slope of the secant?

Steps:

Determine where IROC $=$ AROC.

$v(t)=s'(t)=3t^{2}-10t+4=\frac{s(5)-s(0)}{5-0}$

real velocity $=$ average velocity
slope of tangent $=$ slope of secant
Solve for $t$ to find out WHERE the equation above holds true.

The Mean Value Theorem

The Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$ , then there exists at least one point $x = c$ in $\left(a, b\right)$ such that $\frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) d x= f(c)$ .

First quadrant, bell curve labeled, f of x, left end point on the y axis, labeled, a, right end point labeled, b, dashed horizontal segment, about 1 fourth up from x axis to peak, labeled average, & shaded rectangle between, A & b, segment & x axis, & circled points where the curve intersects the dashed segment, corresponding to tick marks on x axis, labeled, c subscript 1 & c subscript 2.

The Mean Value Theorem for Derivatives

If $F$ is continuous on $[a, b]$ and differentiable on $\left(a, b\right)$ , then there exists at least one point $x = c$ in $\left(a, b\right)$ such that $F ^ { \prime } ( c ) = \frac { F ( b ) - F ( a ) } { b - a } = f ( c )$ .