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

Textbook Question

Chapter 3, 3.4 - Problem 39 - Calculus of a Single Variable (10th Edition, Ron Larson).
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thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on

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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loves2learn | (Level 3) Salutatorian

Posted on

Find the first derivative.

Given,

`f(x)=x+4/x `

Then,

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

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=1-4/x^2 `

`x=+-2 ` (critical points)

These will be a relative extrema if it changes sign, so find 2 values around both to test using the first derivative test, like -3 and -1 for the first  and 1 and 3 for the second.

`f'(-3)=5/9 `

`f'(-1)=-3 `

Therefore, this a relative extrema. The complete point is `(2 ,4 )`

Do the same thing for the other critical point

`f'(1)=-3 `

`f'(3)=5/9 `

Therefore, this a relative extrema as well. The complete point is `( -2, -4)`

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