Question

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)

$f(x)=x^{4}-8x^{3}+7$

relative maximum $(x,y)=(quad)$

relative minimum $(x,y)=(quad)$

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=x^{4}-8x^{3}+7\), we have \(f^\prime(x)=4x^{3}-24x^{2}=4x^{2}(x - 6)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(4x^{2}(x - 6)=0\).
Solving \(4x^{2}(x - 6)=0\) gives \(x = 0\) or \(x=6\)

Step3: Find the second derivative

Differentiate \(f^\prime(x)=4x^{3}-24x^{2}\) using the power rule. \(f^{\prime\prime}(x)=12x^{2}-48x=12x(x - 4)\)

Step4: Apply the second - derivative test

• For \(x = 0\):

Substitute \(x = 0\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(0)=12\times0\times(0 - 4)=0\). The second - derivative test is inconclusive.
We can use the first - derivative test.
For \(x\lt0\), say \(x=-1\), \(f^\prime(-1)=4\times(-1)^{2}\times(-1 - 6)=4\times1\times(-7)=-28\lt0\)
For \(0\lt x\lt6\), say \(x = 1\), \(f^\prime(1)=4\times1^{2}\times(1 - 6)=4\times1\times(-5)=-20\lt0\)
Since the sign of \(f^\prime(x)\) does not change around \(x = 0\), there is no relative extremum at \(x = 0\)

• For \(x = 6\):

Substitute \(x = 6\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(6)=12\times6\times(6 - 4)=12\times6\times2 = 144\gt0\)
When \(x = 6\), \(f(6)=6^{4}-8\times6^{3}+7=1296-1728 + 7=-425\)

Answer:

relative maximum \((x,y)=\text{DNE}\)
relative minimum \((x,y)=(6,-425)\)