# Rate of Change Questions? [closed]

If $h(t)$ represents the height of an object above ground level at time $t$ and $h(t)$ is given by $h(t)=-16t^2+13t+1$ find the height of the object at the time when the speed is zero.

Suppose $h(t)=t^2+14t+7$ . Find the instantaneous rate of change of $h(t)$ with respect to $t$ at $t=2$ .

Suppose $G(x)=6x^2+x+4$ . Find a number $b$ such that $G'(b)=7$ .

Let $g(x)=2x^2+4x+1$ . Find a value of $c$ between 1 and 3 such that the average rate of change of $g(x)$ from $x=1$ to $x=3$ is equal to the instantaneous rate of $g(x)$ at $x=c$ .

Let $F(s)=5s^2+3s+4$ . Find a value of $d$ greater than $0$ such that the average rate of change of $F(s)$ from $0$ to $d$ equals the instantaneous rate of change of $F(s)$ at $s=1$.

Let $f(x)=x^2+x+13$. What is the value of $x$ for which the tangent line to the graph of $y=f(x)$ is parallel to the $x$-axis?

WHAT DO I DO?

## closed as off-topic by colormegone, Pragabhava, Davide Giraudo, Mark Viola, HirshyDec 14 '15 at 18:28

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• Too many questions. – callculus Dec 14 '15 at 13:36

Before starting any calculations I would first recognize the following:

$h(t)=-16t^2+13t+1$

$v(t)=h'(t)=-32t+13=0$

We know that $t=13/32$ when speed is $0$.

$h(13/32) = -16(13/32)^2 +13(13/32) +1 = 3.6406$

$h(t)=t^2+14t+7$

$h'(t)=2t+14$

$h'(2) = 2(2)+14 = 18$ : instantaneous rate of change of $h(t)$ with respect to $t$ at $t=2$ .

$G(x) = 6x^2+x+4$ $G'(x) = 12x+1$ $G'(b) =12b+1 = 0$ $\rightarrow$ $b=-1/12$

$g(x)=2x^2+4x+1$

$g(3)=2(3)^2+4(3)+1 =31$

$g(1)=2(1)^2+4(1)+1 = 7$

Average rate of change of $g(x)$ from $x=1$ to $x=3$ is given by $\frac{g(3)-g(1)}{(3-1)} =(31-7)/2 = 12$ Average rate of change is $12$ Instantaneous rate of change is $4x+4$ Instantaneous rate of change at $x=c$ is $4c + 4$

$4c + 4=12$

$4c=8$

$c=2$

Average rate of change of $F(s)$ from $0$ to $d = \frac{F(d)-F(0)}{d}$

$F(s) =5s^2+3s+4$

$F(d) =5d^2+3d+4$

$F(0)=4$

$\frac{F(d)-F(0)}{d} = \frac{(5d^2+3d+4)-4}{d} = 5d+3$

Instantaneous rate of change of $F(s)$ at $s=1$. $F'(s) = 10s+3$

$F'(1)=13$

$5d+3=13$

$5d=10$

$d=2$

Lastly, When the Tangent line to $y=f(x)$ is: $f'(x) =2x+1$.

If is is parallel to the x-axis, $2x+1 = 0$ (slope is $0$)

$2x+1=0$

$x=-1/2$

It is crucial to keep track of variables when solving rate of change problems!

• It is also crucial to learn how to format your posts using LaTeX. – Alex M. Dec 14 '15 at 13:51