SOLUTION: I am having a hard time with multiplying out (2+h)^3. I have tried to do
(2+h)(2+h)(2+h) and did (2^2+2h+h^2)= (4+2h+h^2) . I am using this in Derivatives.
So If f(x)=x^3 I
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-> SOLUTION: I am having a hard time with multiplying out (2+h)^3. I have tried to do
(2+h)(2+h)(2+h) and did (2^2+2h+h^2)= (4+2h+h^2) . I am using this in Derivatives.
So If f(x)=x^3 I
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Question 494627: I am having a hard time with multiplying out (2+h)^3. I have tried to do
(2+h)(2+h)(2+h) and did (2^2+2h+h^2)= (4+2h+h^2) . I am using this in Derivatives.
So If f(x)=x^3 I have to find the tangent line of the equation through point (2,8).
So F'(2+h)= ((2+h))^3 Which is (4+2h+h^2)
Then F'(2)= (2)^3 Which is (8)
So then I take (4+2h+h^2)-(8)
and I get -4+2h+h^2/h Do you see my problem.
I have tried this so many different ways. Where am I making my error? Found 2 solutions by richard1234, MathLover1:Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! You should definitely review how to multiply polynomials before pursuing a calculus course. To multiply out (2+h)^3, use the binomial theorem and expand.
This is assuming you want to evaluate where x=2.
Or, if you know that the derivative of x^3 is 3x^2, we can easily find the tangent line.
Hence we can write the tangent line in point-slope form
