determine whether the given critical point is the location of a maximum, a minimum, or a point of…

determine whether the given critical point is the location of a maximum, a minimum, or a point of inflection. make a table and show your work.\na. $y = x^{5}-2x^{3}-2x^{2},x = 0$\nb. $f(x)=2x^{3}+4x^{2}-2x - 3,x=-1$

determine whether the given critical point is the location of a maximum, a minimum, or a point of inflection. make a table and show your work.\na. $y = x^{5}-2x^{3}-2x^{2},x = 0$\nb. $f(x)=2x^{3}+4x^{2}-2x - 3,x=-1$

Answer

Explanation:

Step1: Find the first - derivative

For $y = x^{5}-2x^{3}-2x^{2}$, using the power rule $(x^n)'=nx^{n - 1}$, we have $y'=5x^{4}-6x^{2}-4x$.

Step2: Find the second - derivative

Differentiate $y'$ with respect to $x$. $y'' = 20x^{3}-12x - 4$.

Step3: Evaluate the second - derivative at the critical point $x = 0$

Substitute $x = 0$ into $y''$. $y''(0)=20(0)^{3}-12(0)-4=-4<0$. Since $y''(0)<0$, the function has a local maximum at $x = 0$.

For $f(x)=2x^{3}+4x^{2}-2x - 3$,

Step1: Find the first - derivative

Using the power rule, $f'(x)=6x^{2}+8x - 2$.

Step2: Find the second - derivative

Differentiate $f'(x)$ with respect to $x$. $f''(x)=12x + 8$.

Step3: Evaluate the second - derivative at the critical point $x=-1$

Substitute $x=-1$ into $f''(x)$. $f''(-1)=12(-1)+8=-4<0$. Since $f''(-1)<0$, the function has a local maximum at $x=-1$.

Answer:

a. The function $y = x^{5}-2x^{3}-2x^{2}$ has a local maximum at $x = 0$. b. The function $f(x)=2x^{3}+4x^{2}-2x - 3$ has a local maximum at $x=-1$.