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

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

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

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

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

MertsjBest ResponseYou've already chosen the best response.0
How is anyone supposed to help you when we have no idea what the functions are?
 one year ago

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

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

tcarroll010Best ResponseYou've already chosen the best response.3
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 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.
 one year ago

onegirlBest ResponseYou've already chosen the best response.0
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)
 one year ago

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

tcarroll010Best ResponseYou've already chosen the best response.3
his is just what I said also. Exactly what I said.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.3
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.
 one year ago

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

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

tcarroll010Best ResponseYou've already chosen the best response.3
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.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.3
This 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.
 one year ago

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

tcarroll010Best ResponseYou've already chosen the best response.3
If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.3
Good luck to you in all of your studies and thx for the recognition! @onegirl And you're welcome!
 one year ago
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