CPM Homework Help : CALC3RD Problem 7-8

For each function below, calculate the average value over the given interval and state the value of $t$ such that $g\left(t\right)$ equals the average value.

$g(t) = 3t + 6 \text{ for } [0, 8]$
Hint (a):
Read the Math Note about how to compute the Mean Value of $g\left(x\right)$ , given $g\left(x\right)$ .
Hint (a):
To find the the time the function is at its average value, let $g\left(t\right) =$ the average value and solve for $t$ .

$g(t) = 3e^t \text{ for } [0, 1]$
Partial Answer (b):
Average Value $= 3e − 3$ . Now find the time, $t$ , that $g\left(t\right) =$ its average value.

Hint (b):

Average (Mean) Values

To calculate the mean (average) value of a finite set of items, add up the values of items and divide by the number of items.

Integrals help us add over a continuous interval. Therefore, for any continuous function $f$ :

$\frac{\int_a^bf(x)dx}{b-a}=$ mean value of $f$ over $[a, b]$

Since $\int _ { a } ^ { b } f ( x ) d x$ , we can also calculate the average value of any function $f$ using its antiderivative $F$ . Its average slope gives the average rate of change of $F$ , which is the same as the average value of $f.$

$\frac{\int_a^bf(x)dx}{b-a}=\frac { F ( b ) - F ( a ) } { b - a }=$ mean rate of change of $F ′$ over $[a, b]$

First quadrant, 2 tick marks on x axis, first at the origin labeled, A, second almost to the right end, labeled b, Increasing curve labeled, capital F of x, starting at the origin, changing from concave up to concave down, in center of quadrant, ending at point corresponding to, b, almost at the top, with dashed segment labeled, m = average, from origin to end point of curve.

The Mean Value Theorem states that a differentiable function will reach its average (mean) value at least once on any closed interval.
Check your values of t in parts (a) and (b). Are they within the the given closed intervals?

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 $F ^ { \prime } ( c ) = \frac { F ( b ) - F ( a ) } { b - a } = f ( c )$ .