How do use the first derivative test to determine the local extrema #f(x)= -x^3 + 12x#?

Question

Samantha Adkison · Accepted Answer

Refer to explanation

To find the local extrema we need the zeroes of the derivative of f Hence we have that #f'(x)=0=>-3x^2+12=0=>-x^2+4=0=>x=+-2# Now we see how f'(x) behaves around the zeroes graph{-x^2+4 [-10, 10, -5, 5]} We see at x=-2 f'(x)>0 and at x=2 f'(x)<0 Hence at x=2 local maxima f(2)=16 and at x=-2 local minima f(-2)=-16 graph{-x^3+12x [-40, 40, -20, 20]}

Paisley Johnson · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = -x^3 + 12x $: 1. Find the first derivative of the function: $ f'(x) = -3x^2 + 12 $. 2. Set $ f'(x) $ equal to zero and solve for critical points: $ -3x^2 + 12 = 0 $. $ x^2 = 4 $. $ x = \pm 2 $. 3. Determine the sign of the first derivative in the intervals created by the critical points ($ x = -2 $, $ x = 2 $) by testing values within these intervals. - For $ x < -2 $, choose $ x = -3 $: $ f'(-3) = -3(-3)^2 + 12 = -3(9) + 12 = -27 + 12 = -15 < 0 $. - For $ -2 < x < 2 $, choose $ x = 0 $: $ f'(0) = -3(0)^2 + 12 = 12 > 0 $. - For $ x > 2 $, choose $ x = 3 $: $ f'(3) = -3(3)^2 + 12 = -3(9) + 12 = -27 + 12 = -15 < 0 $. 4. Analyze the signs of $ f'(x) $ to determine the nature of the critical points: - At $ x = -2 $, $ f'(x) $ changes from negative to positive, indicating a local minimum. - At $ x = 2 $, $ f'(x) $ changes from positive to negative, indicating a local maximum. Therefore, $ f(x) = -x^3 + 12x $ has a local minimum at $ x = -2 $ and a local maximum at $ x = 2 $.