How do you find and classify local maxima, local minima, and all critical points of #f(x) = x^3+x^2-4x-4#?

2 Answers
Jul 27, 2018

Answer:

There is a local maximum at #(-1.535, ).879# and a local minimum at #(0.869, -6.065)#. The point of inflection is #=(-1/3, -2.593)#.

Explanation:

The function is

#f(x)=x^3+x^2-4x-4#

As this is a polynomial function, the domain is #RR#

Calculate the first derivative

#f'(x)=3x^2+2x-4#

The critical points are when

#f'(x)=0#

That is

#3x^2+2x-4=0#

The solutions to this quadratic equation are

#x=(-2+-sqrt(4+48))/(2*3)=(-2+-sqrt52)/6#

Therefore,

#x_1=(-2-sqrt(52))/6#

#x_2=(-2+sqrt(52))/6#

To determine the nature of thr critical points, you can build either variation chart or calculate the second derivatives.

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##x_2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaaa)##↗#

There is a local maximum at #(-1.535, ).879# and a local minimum at #(0.869, -6.065)#

Calculate the second derivative

#f''(x)=6x+2#

And the point of inflections when

#f''(x)=0#

#=>#, #x=-1/3#

The point of inflection is #=(-1/3, -2.593)#

graph{x^3+x^2-4x-4 [-10, 10, -5, 5]}

Answer:

Point of maximum: #x=-1.535#

Point of minimum: #x=0.868#

Point of inflection: #x=-1/3#

Explanation:

Given function:

#f(x)=x^3+x^2-4x-4#

#f'(x)=3x^2+2x-4#

#f''(x)=6x+2#

For maximum or minimum points, we must have

#f'(x)=0#

#\therefore 3x^2+2x-4=0#

#x=\frac{-2\pm\sqrt{2^2-4(3)(-4)}}{2(3)}#

#x=\frac{-1\pm\sqrt{13}}{3}#

#=0.868 ,-1.535#

#\implies f''(0.868)=6(0.868)+2=7.208>0#

hence, given function is minimum at #x=0.868#

#\implies f''(-1.535)=6(-1.535)+2=-7.21<0#

hence, given function is maximum at #x=-1.535#

Now, the point of inflection will occur at #f''(x)=0#

#\therefore 6x+2=0#

#x=-1/3#