Derivative Tests?

Question

Take the function f(x) = 9 x + 3 x^−1

Local minimum at x equals ___ with value___
Local maximum at x equals ___ with value ___

Answer 1

First, find the derivative and set it equal to 0 to find the critical points.

f'(x) = 9 - 3x^-2
9 - 3x^-2 = 0
-3x^-2 = -9
x^-2 = 3
1/(x^2) = 3
x^2 = 1/3
x = ±1/sqrt(3)

Now use the second derivative to test the critical points:
f''(x) = 6x^-3

At x = -1/sqrt(3)
f'' = 6(-1/sqrt(3))^-3, which is negative
So there is a local maximum at x = -1/sqrt(3)

At x = 1/sqrt(3)
f'' = 6(1/sqrt(3))^-3, which is positive
So there is a local minimum at x = 1/sqrt(3)

To find the values at those points, plug them into the original function, f(x).

Answer 2

y' = 9 -3/x^2 = 0
9x^2=3
x^2=1/3
x = +/- sqrt(1/3) = +/- sqrt(3)/3
Local minimum at x equals sqrt(3)/3 with value f(sqrt(3)/3)
Local maximum at x equals -sqrt(3)/3 with value f(-sqrt(3)/3)

Answer 3

yes