# `f(x) = x + 4/x` Find all relative extrema, use the second derivative test where applicable. First derivative,

`f'(x) = 1+4*(-1)*x^(-1-1) = 1-4x^-2`

At critical points, `f'(x) = 0,`

`1-4/x^2 = 0`

`1=4/x^2`

`x^2=4 `

`x = -2 or x = +2`

At x = -2, f(x) = -2+4/(-2) = -4.

At x = 2, f(x) = 2+4/2 = 4

Therefore, two points (-2,-4) and (2,4) ...

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First derivative,

`f'(x) = 1+4*(-1)*x^(-1-1) = 1-4x^-2`

At critical points, `f'(x) = 0,`

`1-4/x^2 = 0`

`1=4/x^2`

`x^2=4 `

`x = -2 or x = +2`

At x = -2, f(x) = -2+4/(-2) = -4.

At x = 2, f(x) = 2+4/2 = 4

Therefore, two points (-2,-4) and (2,4)  are critical points.

Second derivative tes

`f''(x) = 0-4*(-2)*x^(-2-1) = 8x^-3`

At x = -2, f''(x) = 8*(-2)^-3 = -1 < 0 This suggests that (-2,-4) is a local maximum.

At x=2, f''(X) = 8*(2)^-3 = 1 > 0 This suggests that (2,4) is a local minimum.

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