Question

the function $f(x) = -2x^3 + 36x^2 - 162x + 7$ has one local minimum and one local maximum. this function has a local minimum at $x = \square$ with value $\square$ and a local maximum at $x = \square$ with value $\square$ question help: video message instructor

Explanation:

Step1: Find the first derivative

To find local extrema, we first find the derivative of \( f(x)=-2x^{3}+36x^{2}-162x + 7 \). Using the power rule, the derivative \( f'(x) \) is:
\( f'(x)=-6x^{2}+72x - 162 \)

Step2: Find critical points

Set \( f'(x) = 0 \) to find critical points:
\( -6x^{2}+72x - 162=0 \)
Divide both sides by -6:
\( x^{2}-12x + 27 = 0 \)
Factor the quadratic:
\( (x - 3)(x - 9)=0 \)
So the critical points are \( x = 3 \) and \( x = 9 \)

Step3: Find the second derivative

To determine if these are local minima or maxima, we find the second derivative \( f''(x) \):
\( f''(x)=-12x + 72 \)

Step4: Classify critical points

For \( x = 3 \):
\( f''(3)=-12(3)+72=-36 + 72 = 36>0 \), so \( x = 3 \) is a local minimum.
For \( x = 9 \):
\( f''(9)=-12(9)+72=-108 + 72=-36<0 \), so \( x = 9 \) is a local maximum.

Step5: Find the function values at critical points

For \( x = 3 \):
\( f(3)=-2(3)^{3}+36(3)^{2}-162(3)+7=-2(27)+36(9)-486 + 7=-54 + 324-486 + 7=-209 \)
For \( x = 9 \):
\( f(9)=-2(9)^{3}+36(9)^{2}-162(9)+7=-2(729)+36(81)-1458 + 7=-1458+2916-1458 + 7=7 \)

Answer:

This function has a local minimum at \( x = 3 \) with value \( -209 \) and a local maximum at \( x = 9 \) with value \( 7 \)