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

1 Answer
Jan 14, 2018

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]}