Question

Write the equation (estimated) of the tangent line.h(x) = f(x)g(x) at x = 1.

Answer 1

@rizwan_uet can u help?

Answer 2

Would you please post the problem correctly?

Answer 3

okay

Answer 4

\[h(x) = f(x)g(x) \] at x = 1

Answer 5

@tcarroll010 can u help?

Answer 6

@RadEn can u help?

Answer 7

How is anyone supposed to help you when we have no idea what the functions are?

Answer 8

well thats the question, it says find an equation of a tangent line to h(x) = f(x)g(x) at x = 1

Answer 9

there are no functions..

Answer 10

so can anyone help?

Answer 11

You can get the first derivative of h(x) as follows with the product rule:

h'(x) = f'(x)g(x) + f(x)g'(x)

That will be your slope at any point where the first derivative is defined. Now, you use the point-slope formula for the equation of a line:\[y - y _{1} = m(x - x _{1})\]and here, "m" is as defined above with the first derivative for your slope. Make sure you substitute "1" where you see "x" in that first equation. x1 is 1 and y1 is h(1). So, just make the substitutions and that is the equation for the tangent line.

Answer 12

I did this but my teacher told me to write the equation an estimated one "the curve is (1, h(1)) , using the product rule h'(x) = f'(x) g'(x) = + g(x)f'(x) substituting x= 1 into the 1st derivative the slope will be h'(1) = f(1) g'(1) + g(1)f'(1) so the slope h'(1) and the point (1,h(1) so y = (f'(1)g(1) +f(1)g'(1) (x - 1) + f(1)g(1)

Answer 13

so first i have to get the derivative of h'(x) f'(x) g(x) + f(x)g'(x) right?

Answer 14

his is just what I said also. Exactly what I said.

Answer 15

okay

Answer 16

The only difference, which is really nothing, is that I said y1 is h(1). That is the same as f(1)g(1). Just written slightly differently, that's all. And that's all you have to do. And you put the h(1) on the right which is fine. Same answer.

Answer 17

Well i did this but my teacher said i need to write an equation and estimated one. :/

Answer 18

so maybe my final answer needs to be estimated?

Answer 19

Yours and my answer, which is the same and is the correct and exact answer. Let me see if we can estimate this for x = 1. Hold on a minute.

Answer 20

okay

Answer 21

This can be re-written:

y = [f'(1)g(1) + f(1)g'(1)](x - 1) + f(1)g(1)

y = [f'(1)g(1) + f(1)g'(1)]x + ( f(1)g(1) - [f'(1)g(1) + f(1)g'(1)] )

Now, if the derivative at x = 1 is small, we could drop it out of the far right side of the above equation as it is in the form:

y = mx + b

y = [f'(1)g(1) + f(1)g'(1)]x + f(1)g(1)

But that is if the derivative at x = 1 is small. Then, we have a good approximation.

Answer 22

okay

Answer 23

well i'll try this thanks

Answer 24

If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.

Answer 25

okay

Answer 26

Good luck to you in all of your studies and thx for the recognition! @onegirl

And you're welcome!