#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]}
