We are given:
[tex]f(x)=x^3+5x^2-7x-3[/tex]Now, we know that in order to determine the critical points we derivate and the derivative is then equal to 0, that is:
[tex]f^{\prime}(x)=3x^2-10x-7=0[/tex]Now, we solve for x, that is:
[tex]3x^2+10x-7=0\Rightarrow x=\frac{-(10)\pm\sqrt[]{(10)^2-4(3)(-7)}}{2(3)}[/tex][tex]\Rightarrow\begin{cases}x=-\frac{5+\sqrt[]{46}}{3}\Rightarrow x\approx-3.9 \\ \\ x=\frac{-5+\sqrt[]{46}}{3}\Rightarrow x\approx0.6\end{cases}[/tex]So, the critical points of the function are:
[tex]\begin{cases}x=-\frac{5+\sqrt[]{46}}{3} \\ \\ x=\frac{-5+\sqrt[]{46}}{3}\end{cases}[/tex]Now, we determine the y-components of the points, that is:
[tex]\begin{cases}f(-\frac{5+\sqrt[]{46}}{3})=(-\frac{5+\sqrt[]{46}}{3})^3+5(-\frac{5+\sqrt[]{46}}{3})^2-7(-\frac{5+\sqrt[]{46}}{3})-3\Rightarrow f(-\frac{5+\sqrt[]{46}}{3})=41.03608735 \\ \\ f(\frac{-5+\sqrt[]{46}}{3})=(\frac{-5+\sqrt[]{46}}{3})^3+5(\frac{-5+\sqrt[]{46}}{3})^2-7(\frac{-5+\sqrt[]{46}}{3})-3\Rightarrow f(\frac{-5+\sqrt[]{46}}{3})=-5.184235498\end{cases}[/tex]So, the two critical points are:
[tex](-\frac{5+\sqrt[]{46}}{3},41.03608735)[/tex]and:
[tex](\frac{-5+\sqrt[]{46}}{3},-5.184235498)[/tex]This can be seing as follows:
