Use the second derivative test to find the local extrema of the given function.
$g(x)=x^3-6x^2+9x+2$

Local max of $6$ at $x=1$ , local min of 2 at $x=3$

Local min of $6$ at $x=1$ , local max of 2 at $x=3$

Local max of $6$ at $x=-1$ and $x=1$ , local min of $-6$ at $x=-3$ and $x=3$

No local extrema

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Step 1: Begin by finding the first derivative of the function g(x). The given function is g(x) = x^3 - 6x^2 + 9x + 2. Use the power rule to differentiate term by term: g'(x) = 3x^2 - 12x + 9.

Step 2: Set the first derivative g'(x) equal to zero to find the critical points. Solve the equation 3x^2 - 12x + 9 = 0. Factorize or use the quadratic formula to find the values of x.

Step 3: Once the critical points are determined, find the second derivative of the function g(x). Differentiate g'(x) = 3x^2 - 12x + 9 to get g''(x) = 6x - 12.

Step 4: Use the second derivative test to classify the critical points. Substitute each critical point into g''(x). If g''(x) > 0 at a critical point, it is a local minimum. If g''(x) < 0, it is a local maximum. If g''(x) = 0, the test is inconclusive.

Step 5: Finally, substitute the critical points into the original function g(x) to find the corresponding function values. These values, along with the classification from the second derivative test, will give the local extrema of the function.

5:34m

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Use the second derivative test to find the local extrema of the given function.g(x)=x3−6x2+9x+2g(x)=x^3-6x^2+9x+2g(x)=x3−6x2+9x+2

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Use the second derivative test to find the local extrema of the given function.
$g(x)=x^3-6x^2+9x+2$