What are the local extrema of #f(x)=(x-3)(x^2-2x-5)#?

1 Answer
Jan 14, 2018

Answer:

#f(x)# has a local maximum at #approx(0.1032, 15.0510)#
#f(x)# has a local minimum at #approx(3.2301, -0.2362)#

Explanation:

#f(x) = (x-3)(x^2-2x-5)#

Apply product rule.

#f'(x) = (x-3) * d/dx (x^2-2x-5) + d/dx(x-3)* (x^2-2x-5)#

Apply power rule.

#f'(x) = (x-3)(2x-2) + 1*(x^2-2x-5)#

#= 2x^2-8x+6 + x^2-2x-5#

#= 3x^2-10x+1#

For local extrema #f'(x)=0#

Hence, #3x^2-10x+1 =0#

Apply Quadratic Formula.

#x= (+10+-sqrt((-10)^2-4*3*1))/(2*3)#

#= (10+-sqrt(88))/6#

# approx 3.2301 or 0.1032#

#f''(x) = 6x-10#

For local maximum #f'' <0# at extreme point.

For local minimum #f'' >0# at extreme point.

Testing #f''(3.2301)>0 ->f(3.2301) = f_min#

Testing #f''(0.1032)<0 ->f(0.1032) = f_max#

Hence, #f_max approx (0.1032-3)(0.1032^2-2*0.1032-5)#

#approx 15.0510#

And, #f_min approx (3.2301-3)(3.2301^2-2*3.2301-5)#

#approx -0.2362#

#:. f(x)# has a local maximum at #approx(0.1032, 15.0510)#

#and f(x)# has a local minimum at #approx(3.2301, -0.2362)#

We can see these local extrema by zooming to the relevant points on the graph of #f(x)# below.

graph{(x-3)(x^2-2x-5) [-29.02, 28.72, -6.2, 22.63]}