Question 560133: F(x) = 4x^3 - 5x 2x^(1/2). Determine the slope of the tangent when x = 4?
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! This is a calculus 1 question.
You need to find the derivative of F(x).
I'm looking for F'(x). So, I will do this by differenting termwise.
(4x^3)' = 12x^2
(-5x)' = -5
(2x^1/2)' = x^(-1/2)
We can write x^(-1/2) as 1/sqrt{x}.
So, F'(x) = 12x^2 + 1/sqrt{x} - 5
To find the slope of the tangent line, we now replace every x you see with 4 and simplify.
F'(4) = 12(4)^2 + 1/sqrt{4} - 5
F'(4) = 12*16 + 1/2 - 5
F'(4) = 375/2
The slope of the tangent line is 375/2.
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