Question

Find the net change in the value of function $h\left(t\right)=t^2+5$for the inputs from $t=-3\mathrm{to}t=6$.

Short answer

The net change in the value of function is $h\left(6\right)-h\left(-3\right)=27$.

Step-by-step solution

Step 1. Evaluate the value of function at extreme points.

From the given function,

$h\left(t\right)=t^2+5....\left(1\right)$

Evaluate $h\left(-3\right)\text{and}h\left(6\right)$gives,

role="math" localid="1644531344415" $$\begin{gathered} h\left(-3\right)={\left(-3\right)}^2+5....\left(\text{Substitute}t=-3\text{in}\left(1\right)\right) \\ =14 \\ h\left(6\right)={\left(6\right)}^2+5....\left(\text{Substitute}t=6\text{in}\left(1\right)\right) \\ =41 \end{gathered}$$

Step 2. Obtain difference at extreme points.

Evaluate difference at extreme points gives the net change $f\left(b\right)-f\left(a\right)$where $b>a$. Therefore,

role="math" localid="1644531411772" $\begin{array}{c}h\left(6\right)-h\left(-3\right)=41-14\\ =27\end{array}$

Step 3. Description of steps.

The net change in the value of function is $h\left(6\right)-h\left(-3\right)=27$.