How do you find the critical numbers for #f(x)=x^2-6x# to determine the maximum and minimum?

Question

William Abdalla · Accepted Answer

Differentiate to find the first derivative which will be zero at a max or min.

#f(x)=x^2-6x# Differentiating gives: #f'(x)=2x-6# At a max or min, #f'(x)=0 => 2x-6=0# # :. 2x=6=>x=3 # So there is one critical point (could be max or min; in fact it is a minimum by observation as its is a +ve U shaped quadratic) when #x=3#

Charles Arocha · Answer

undefined To find the critical numbers for $ f(x) = x^2 - 6x $, we first find the derivative, $ f'(x) $, and then solve for $ x $ where $ f'(x) = 0 $ or where $ f'(x) $ is undefined.

The derivative of $ f(x) = x^2 - 6x $ is $ f'(x) = 2x - 6 $.

Setting $ f'(x) = 0 $, we have $ 2x - 6 = 0 $. Solving for $ x $, we get $ x = 3 $.

So, the critical number for $ f(x) $ is $ x = 3 $.

To determine whether $ f(x) $ has a maximum or minimum at $ x = 3 $, we can use the first or second derivative test.

Considering the second derivative test:

If $ f''(3) > 0 $, then $ f(x) $ has a local minimum at $ x = 3 $.
If $ f''(3) < 0 $, then $ f(x) $ has a local maximum at $ x = 3 $.
If $ f''(3) = 0 $, the test is inconclusive.

The second derivative of $ f(x) = x^2 - 6x $ is $ f''(x) = 2 $.

Evaluating $ f''(3) $, we get $ f''(3) = 2 $.

Since $ f''(3) > 0 $, $ f(x) $ has a local minimum at $ x = 3 $.