Question

a function ( f(x) ) is given. find the ( x ) values where ( f(x) ) has a relative maximum or minimum. (note: these are the same functions as in exercise group 15 - 28.)
( f(x)=-x^{4}+62 x^{2}+120 x + 4 )
enter the points where ( f(x) ) has a relative maximum, separating with commas if needed. if there are none, enter none.
enter the points where ( f(x) ) has a relative minimum, separating with commas if needed. if there are none, enter none.

Explanation:

Step1: Find the first - derivative of \(f(x)\)

Given \(f(x)=-x^{4}+62x^{2}+120x + 4\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-4x^{3}+124x + 120\).
Factor out \(-4\): \(f^\prime(x)=-4(x^{3}-31x - 30)\).
By trial - and - error (using the rational root theorem, if \(x = -1\), then \((-1)^{3}-31(-1)-30=-1 + 31-30 = 0\)).
So \(x^{3}-31x - 30=(x + 1)(x^{2}-x - 30)\).
Factor \(x^{2}-x - 30=(x - 6)(x+5)\).
So \(f^\prime(x)=-4(x + 1)(x - 6)(x + 5)\).

Step2: Find the second - derivative of \(f(x)\)

Using the product rule \((uvw)^\prime=u^\prime vw+uv^\prime w+uvw^\prime\) (where \(u=-4\), \(v=x + 1\), \(w=(x - 6)(x + 5)=x^{2}-x - 30\)).
\(f^{\prime\prime}(x)=-4[(x - 6)(x + 5)+(x + 1)(2x - 1)+(x + 1)(x - 6)]\).
Expand:

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Step3: Use the second - derivative test

Evaluate \(f^{\prime\prime}(x)\) at the critical points \(x=-5,x=-1,x = 6\).

• For \(x=-5\):

\(f^{\prime\prime}(-5)=-4(4\times(-5)^{2}-5\times(-5)-37)=-4(100 + 25-37)=-4\times88=-352\lt0\). So \(x=-5\) is a relative maximum.

• For \(x=-1\):

\(f^{\prime\prime}(-1)=-4(4\times(-1)^{2}-5\times(-1)-37)=-4(4 + 5-37)=-4\times(-28)=112\gt0\). So \(x=-1\) is a relative minimum.

• For \(x = 6\):

\(f^{\prime\prime}(6)=-4(4\times6^{2}-5\times6-37)=-4(144-30 - 37)=-4\times77=-308\lt0\). So \(x = 6\) is a relative maximum.

Answer:

• Relative maximum: \(-5,6\)
• Relative minimum: \(-1\)