Question

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.
a. (y = x^{5}-2x^{3}-2x^{2},x = 0)
b. (f(x)=2x^{5}+4x^{2}-2x - 3,x=-1)

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 \(x=-1\)