PLEASE HELP!! If f (x) = 4x^-2 + 1/4x^2 + 4, then f’(2) =
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Work Shown:
f(x) = 4x^(-2) + (1/4)x^2 + 4
f ' (x) = 4*(-2)x^(-3) + (1/4)*2x .... power rule
f ' (x) = -8x^(-3) + (1/2)x
f ' (2) = -8(2)^(-3) + (1/2)*(2)
f ' (2) = 0
Finding the derivative, we have that f'(2) = 0, option D.
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[tex](x^n)^{\prime} = nx^{n-1}[/tex]
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The function is:
[tex]f(x) = 4x^{-2} + \frac{x^2}{4} + 4[/tex]
The derivatives are:
[tex](4x^{-2})^{\prime} = -2 \times 4x^{-2 -1} = -8x^{-3}[/tex]
[tex](\frac{x^2}{4})^{\prime} = \frac{2x}{4}[/tex]
[tex](4)^{\prime} = 0[/tex]
Thus, the derivative of the function is:
[tex]f^{\prime}(x) = -8x^{-3} + \frac{2x}{4} = -\frac{8}{x^3} + \frac{2x}{4}[/tex]
At x = 2:
[tex]f^{\prime}(2) = -\frac{8}{2^3} + \frac{2(2)}{4} = -1 + 1 = 0[/tex]
Thus f'(2) = 0, option D.
A similar problem is given at https://brainly.com/question/10037173