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.
^3 = (3C0)2^3 + (3C1)2^2h + (3C2)2h^2 + (3C3)h^3)
This is assuming you want to evaluate  - f(x)}{h})
where x=2.
Or, if you know that the derivative of x^3 is 3x^2, we can easily find the tangent line.
 = 3x^2)
 = 3(2)^2 = 12)
Hence we can write the tangent line in point-slope form
)
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
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