Question

given that

f(x)=x^6h(x)
h(−1)=2
h′(−1)=5

f′(−1)=?

Answer 1

Is it f(x)=x^(6h(x)) or f(x)=(x^6)h(x)?

Answer 2

x^(6)h(x)

Answer 3

Alright, then. We know that f'(x)=h'(x)(x^6)+(6x^5)h(x).

At x=-1, f(-1)=((-1)^6)h(-1)=1*2 (we're given the solution for h(-1)).

We want to find f'(-1)=h'(-1)((-1)^6)+(6(-1)^5)h(-1). From here, we can solve it.

Answer 4

you mind if i ask how?

Answer 5

Plug things in. We know h'(-1), as we're given it. We know h(-1), since we also are given that.

Answer 6

So I am plugging -1 into both equations and doing what with the result? Multiplying?

Answer 7

We want to find f'(-1), right? Taking the derivative of f(x) should give us f'(x). So I took the derivative.

f'(x)=h'(x)(x^6)+(6x^5)h(x)

Now plug in -1 for x. So we have f'(-1)=h'(-1)((-1)^6)+(6(-1)^5)h(-1). By solving the right side, which we can since we know the values for all the terms, we can find f'(-1).

Answer 8

ok let me try

Answer 9

"Do or do not. There is no try." ~ Yoda

Answer 10

;)

Answer 11

For my answer, I received -175

Answer 12

is that correct?

Answer 13

I received -7. At this point it's basic calculation mistakes. You should check your math.

Answer 14

ok thanks

Answer 15

can you help me with this one if you don't mind?

Answer 16

Sure.

Answer 17

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(x). Find each of the following:

u'(1)

v'(1)

Answer 18

Alright, since we're primarily concerned with the functions at 0<x<3, we should just try to model at those points alone. f(x)=x, and g(x)=-x+4. These are evident by looking at the graphs.

u(x)=f(x)g(x)=-(x^2)+4x, u'(x)=-2x+4, u'(1)=2

v(x)=f(x)/g(x)=x/(-x+4), v'(x)=1/(-x+4)+x/(-x+4)^2, v'(1)=(1/3)+(1/9)=4/9

I think my math is right, but you should follow these steps too to make sure.

Answer 19

last one

x 11 1 2 −3
f(x) −1209 1 −3 37
g(x) 2432 2 11 −74
f′(x) −341 −1 −8 −33
g′(x) 683 3 17 67

(fg)′(2)
f(2)/(g(2)+5)
(fg)(1)
(f−g)′(2)

Answer 20

Does it say (fg)'(2) for the first one? I don't think that makes sense. Typo?

Answer 21

You sure it isn't (f(g'(x)) or something?

Answer 22

its (fg)'2

Answer 23

This is a markup I'm unfamiliar with. Sorry, can't help you here.

Answer 24

what about the rest