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^{5}+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)^\prime=nx^{n - 1}), we have (y^\prime=5x^{4}-6x^{2}-4x). For (f(x)=2x^{5}+4x^{2}-2x - 3), (f^\prime(x)=10x^{4}+8x - 2).
Step2: Find the second - derivative
For (y^\prime=5x^{4}-6x^{2}-4x), (y^{\prime\prime}=20x^{3}-12x - 4). For (f^\prime(x)=10x^{4}+8x - 2), (f^{\prime\prime}(x)=40x^{3}+8).
Step3: Evaluate the second - derivative at the critical points
For part a
When (x = 0), (y^{\prime\prime}(0)=20(0)^{3}-12(0)-4=-4<0). Since (y^{\prime\prime}(0)<0), the function (y = x^{5}-2x^{3}-2x^{2}) has a local maximum at (x = 0).
For part b
When (x=-1), (f^{\prime\prime}(-1)=40(-1)^{3}+8=40\times(-1)+8=-32<0). Since (f^{\prime\prime}(-1)<0), the function (f(x)=2x^{5}+4x^{2}-2x - 3) has a local maximum at (x=-1).
Answer:
a. Local maximum at (x = 0) b. Local maximum at (x=-1)