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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@rizwan_uet can u help?

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.0Would you please post the problem correctly?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[h(x) = f(x)g(x) \] at x = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@tcarroll010 can u help?

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.0How is anyone supposed to help you when we have no idea what the functions are?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well thats the question, it says find an equation of a tangent line to h(x) = f(x)g(x) at x = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are no functions..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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 pointslope 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so first i have to get the derivative of h'(x) f'(x) g(x) + f(x)g'(x) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0his is just what I said also. Exactly what I said.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well i did this but my teacher said i need to write an equation and estimated one. :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so maybe my final answer needs to be estimated?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yours 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This can be rewritten: 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well i'll try this thanks

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Good luck to you in all of your studies and thx for the recognition! @onegirl And you're welcome!
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