Answer to Question #291364 in Real Analysis for Dhruv bartwal

Question

Find the local maximum and local minimum values of the function

f(x)= x^3-2x^2-4x+5

Answer 1

Let us find the local maximum and local minimum values of the function $f(x)= x^3-2x^2-4x+5.$

It follows that $f'(x)= 3x^2-4x-4=(x-2)(3x+2).$ Therefore, $f'(x)=0$ implies $x_1=2$ and $x_2=-\frac{2}3.$ Taking into account that $f''(x)= 6x-4,$ we conclude that $f''(2)=8>0, \ f''(-\frac{2}3)=-8<0.$ Consequently, $x_1=2$ is the point of a local minimum and $x_2=-\frac{2}3$ is a the point of local maximum.