Here's the question you clicked on:
onegirl
Write the equation (estimated) of the tangent line.h(x) = f(x)g(x) at x = 1.
@rizwan_uet can u help?
Would you please post the problem correctly?
\[h(x) = f(x)g(x) \] at x = 1
@tcarroll010 can u help?
How is anyone supposed to help you when we have no idea what the functions are?
well thats the question, it says find an equation of a tangent line to h(x) = f(x)g(x) at x = 1
there are no functions..
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.
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)
so first i have to get the derivative of h'(x) f'(x) g(x) + f(x)g'(x) right?
his is just what I said also. Exactly what I said.
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.
Well i did this but my teacher said i need to write an equation and estimated one. :/
so maybe my final answer needs to be estimated?
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.
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.
well i'll try this thanks
If I were the teacher, this is what I'd be looking for. Remember the stipulation of the derivative being small.
Good luck to you in all of your studies and thx for the recognition! @onegirl And you're welcome!