-2
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Pragabhava, Mark Viola, Hirshy
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Too many questions. $\endgroup$ – callculus Dec 14 '15 at 13:36
2
$\begingroup$

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!

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.