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^{3}-6 x^{2}+2 )

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 \(y = f(x)=x^{3}-6x^{2}+2\), the first derivative \(f^\prime(x)=3x^{2}-12x\).
Set \(f^\prime(x) = 0\), so \(3x^{2}-12x=3x(x - 4)=0\).
Solving \(3x(x - 4)=0\) gives \(x = 0\) or \(x = 4\). These are the critical points.

Step2: Find the second derivative

Differentiate \(f^\prime(x)=3x^{2}-12x\) with respect to \(x\). Using the power rule, \(f^{\prime\prime}(x)=6x-12\).

Step3: Apply the second - derivative test

• For \(x = 0\):

Substitute \(x = 0\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(0)=6\times0 - 12=- 12<0\).
When \(x = 0\), \(y=f(0)=0^{3}-6\times0^{2}+2 = 2\). So \((0,2)\) is a relative maximum.

• For \(x = 4\):

Substitute \(x = 4\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(4)=6\times4-12 = 12>0\).
When \(x = 4\), \(y=f(4)=4^{3}-6\times4^{2}+2=64 - 96 + 2=-30\). So \((4,-30)\) is a relative minimum.

Answer:

relative maximum \((x,y)=(0,2)\)
relative minimum \((x,y)=(4,-30)\)