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onegirl
 one year ago
Best ResponseYou've already chosen the best response.0@rizwan_uet can u help?

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

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

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0@tcarroll010 can u help?

Mertsj
 one year 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?

onegirl
 one year 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

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0there are no functions..

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3You 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.

onegirl
 one year 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)

onegirl
 one year 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?

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3his is just what I said also. Exactly what I said.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3The 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.

onegirl
 one year 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. :/

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0so maybe my final answer needs to be estimated?

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3Yours 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.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3This 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.

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0well i'll try this thanks

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.3Good luck to you in all of your studies and thx for the recognition! @onegirl And you're welcome!
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