Question

Based on the attached figure, what is h'(3) ?

please explain? Thanks :)

Answer 1

the figure/graph/info!

Answer 2

Ignore the graph for a moment.
\[\Large\bf\sf h(x)\quad=\quad f(x)^3\]Understand how to find this derivative?\[\Large\bf\sf h'(x)\quad=\quad ?\]

Answer 3

using nx^n-1 ?

Answer 4

Yes, power rule then chain rule.

Answer 5

The graph is actually only used for one small part of the problem :)
So don't let it distract you hehe

Answer 6

okay:)

so what happens now?

do i get

(3)(f(x))^3-1 ?

Answer 7

Don't forget your chain rule!!\[\Large\bf\sf h'(x)\quad=\quad 3f(x)^2\cdot f'(x)\]

Answer 8

Then, evaluate the derivative at x=3.\[\Large\bf\sf h'(3)\quad=\quad 3f(3)^2\cdot f'(3)\]

Answer 9

We can figure out f(3) simply by looking at the graph.
What's the value of the function at x=3 (Where the light blue line crosses 3).

Answer 10

Comeoooooon food +_+ Where you at?

Answer 11

sorry, I lost connection for a moment there...

let me take a look at this :)

Answer 12

so, we're using the graph now right?

Answer 13

yes

Answer 14

if so, at x=3, y=5?

Answer 15

or am i evaluating by plugging in?

Answer 16

\[\Large\bf\sf h'(3)\quad=\quad 3\color{royalblue}{f(3)}^2\cdot f'(3)\]Ok good!\[\Large\bf\sf h'(3)\quad=\quad 3\cdot\color{royalblue}{5}^2\cdot f'(3)\]

Answer 17

So that takes care of the f(3)

Answer 18

:) so now we need to find f ' (3) ?

Answer 19

Try to think about what the derivative represents.

The derivative evaluated at x=3 is
`the slope of the line tangent to the curve at x=3`.

So f'(3) is simply the `slope` of that black line.

Answer 20

*Because it's tangent to the curve at x=3

Answer 21

So we'll go back to our old school slope formula,\[\Large\bf\sf m\quad=\quad \frac{y_2-y_1}{x_2-x_1}\quad=\quad \frac{f(x_2)-f(x_1)}{x_2-x_1}\]

Answer 22

so it would be .1/ .3 ? so 1/3 ?

Answer 23

Mmm yes good!

Answer 24

so does that mean that h'(3) = 1/3 ? or are there more steps?

Answer 25

It means f'(3) = 1/3

Answer 26

Plug it in.

Answer 27

oh yes, so 3 * 5^2 * f ' (3) right?

so 3* 25 * 1/3 = 75/3 = 25 ?

Answer 28

This is our derivative,\[\Large\bf\sf h'(3)\quad=\quad 3f(3)^2\cdot f'(3)\]
And we determined,
\[\Large\bf\sf f(3)=5\]\[\Large\bf\sf f'(3)=\frac{1}{3}\]

Answer 29

Mmm yes, good job!

Answer 30

awesome! thank you :)