How do you find the critical points for #f(x) = x^3 - 15x^2 + 4# and the local max and min?

Question

Ezra Adams · Accepted Answer

The critical points are: #x_1=0# and #x_2=10#.
#x_1# is a maximum, while #x_2# is a minimum.

To identify the critical points, we have to solve the equation: #f'(x)=0# that is: #3x^2-30x=0# #3x(x-10) =0# so that the critical points for #f(x)# are #x_1=0# and #x_2=10#. We can now calculate the second derivative: #f''(x) = 6x-30# and note that around #x_1# #f''(x)# is negative, therefore #x_1# is a maximum, while around #x_2# it is positive, therefore #x_2# is a minimum. graph{x^3-15x^2+4 [-80, 80, -640, 640]}

Adam Adkins · Answer

undefined To find the critical points of $ f(x) = x^3 - 15x^2 + 4 $, we first find its derivative, then set it equal to zero to solve for critical points.

The derivative of $ f(x) $ is $ f'(x) = 3x^2 - 30x $.

Setting $ f'(x) $ equal to zero, we have $ 3x^2 - 30x = 0 $.

Factoring out $ 3x $ gives $ 3x(x - 10) = 0 $.

This yields critical points $ x = 0 $ and $ x = 10 $.

To determine whether these critical points correspond to local maxima, minima, or neither, we can use the second derivative test or analyze the behavior of the function around these points.

Taking the second derivative of $ f(x) $ yields $ f''(x) = 6x - 30 $.

Evaluating $ f''(0) $ and $ f''(10) $:

At $ x = 0 $, $ f''(0) = -30 $, indicating a concave down shape, implying a local maximum.

At $ x = 10 $, $ f''(10) = 30 $, indicating a concave up shape, implying a local minimum.

Therefore, $ f(x) = x^3 - 15x^2 + 4 $ has a local maximum at $ x = 0 $ and a local minimum at $ x = 10 $.