Question

find the x - coordinates of all relative maxima of f(x).
f(x)=-2x^{4}-16x^{3}+23

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y=-2x^{4}-16x^{3}+23\), the first derivative \(f^\prime(x)=-8x^{3}-48x^{2}\).

Step2: Set the first derivative equal to zero

Set \(f^\prime(x)=0\), so \(-8x^{3}-48x^{2}=0\). Factor out \(-8x^{2}\): \(-8x^{2}(x + 6)=0\).
Solving \(-8x^{2}(x + 6)=0\) gives \(x = 0\) or \(x=-6\).

Step3: Find the second derivative

Differentiate \(f^\prime(x)=-8x^{3}-48x^{2}\) using the power rule. The second derivative \(f^{\prime\prime}(x)=-24x^{2}-96x\).

Step4: Test the critical points

• For \(x = 0\): \(f^{\prime\prime}(0)=-24(0)^{2}-96(0)=0\). The second - derivative test is inconclusive.
• For \(x=-6\): \(f^{\prime\prime}(-6)=-24(-6)^{2}-96(-6)=-24\times36 + 576=-864 + 576=-288<0\).

Since \(f^{\prime\prime}(-6)<0\), the function \(f(x)\) has a relative maximum at \(x=-6\).

Answer:

\(-6\)