Respuesta :

Answer: D) 0

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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  • The derivative of an exponent is:

[tex](x^n)^{\prime} = nx^{n-1}[/tex]

  • Additionally, the sum of the derivatives is the derivative of the sum.
  • The derivative of a constant is zero.

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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