Question

Need help understanding how to solve the question! Thank you!!

Answer 1

the answers that are circled are correct, i just dont know how to get them

Answer 2

First, find the equations of the lines f(x) and g(x) in the graph ...

Answer 3

would that be for one point on the graph?

Answer 4

Finding equations is not really necessary.

For \(h(x)=f(x)g(x),~~~h'(x)=f'(x)g(x)+f(x)g'(x) \)

Can you find the value of f'(3) and g'(3) from given graph?

Answer 5

no?!? since the graph is f(x) and g(x)?!?

Answer 6

Yea, you need the equations to find the derivative at f'(3) and g'(3)

Answer 7

Well, basically \(f'(x)\) and \(g'(x)\) is slope of \(f(x)\) and \(g(x)\) respectively.

Knowing equation is not necessary, you can tell rate of change of line from given graph of original function.

Answer 8

Yea, either way will do....

Answer 9

ok, so im confused on how to find f'(x) and g'(x)

Answer 10

At x=3, what is slope of f(x)?

Answer 11

-1

Answer 12

Right, so \(f'(3)=-1\).

Now what's the slope of g(x) at x=3?

Answer 13

Oh so then the slope of g(x)=2 and g'(x)=2....correct?

Answer 14

when x=3

Answer 15

Yeah, so now you have what you need: \(f(3)=2,~g(3)=2,~f'(3)=-1,~g'(3)=2\)

To find \(h'(3)\) just plug in values you have.

Answer 16

\(h'(3) = f'(3)g(3)+f(3)g'(3) = (-1)(2)+(2)(2) = \boxed{2}\)

Answer 17

ok, i understand it now!! Thanks!

Answer 18

Glad we helped.

Answer 19

Do you still need part 2 answered?

Answer 20

if you dont mind, that would be helpful

Answer 21

You agree that \(m'(x) = f'(g(x))g'(x)\), right?

Answer 22

yes, would f(4)=1 f'(4)=-1 g(4)=4 and g'(4)=.5?

Answer 23

g'(4) is actually 2. Because slope at x=4 is 2.

Answer 24

That would give you an answer of -8, correct?

Answer 25

Ok, so you have \(m'(4) = f'(g(4))\cdot g'(4) = f'(4)\cdot 2=-1\cdot2=\boxed{2}\)

Answer 26

Note that \(g(x)\) is nested in \(f'(x)\)

Answer 27

Oops \(\boxed{-2}\)*

Answer 28

oh i multiplied f' by g

Answer 29

So that make sense? Need more clarification?

Answer 30

All good, I get it now! Thank you @geerky42 and @Hyper!!!!!

Answer 31

No problem, mainly thanks to geerky :)