What are the local extrema, if any, of #f (x) = x^3  6x^2  15x + 11 #?
Maxima=19 at x=1
Minimum=89 atx=5
To find the local extrema first find the critical point
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To find the local extrema of the function ( f(x) = x^3  6x^2  15x + 11 ), we first find its derivative and then solve for critical points by setting the derivative equal to zero. After that, we employ the second derivative test to determine the nature of these critical points, whether they correspond to local maxima, local minima, or neither.

Find the derivative of the function: [ f'(x) = 3x^2  12x  15 ]

Set the derivative equal to zero and solve for ( x ) to find critical points: [ 3x^2  12x  15 = 0 ] [ x^2  4x  5 = 0 ]

Use the quadratic formula to solve for ( x ): [ x = \frac{{b \pm \sqrt{{b^2  4ac}}}}{{2a}} ] [ x = \frac{{(4) \pm \sqrt{{(4)^2  4(1)(5)}}}}{{2(1)}} ] [ x = \frac{{4 \pm \sqrt{{16 + 20}}}}{{2}} ] [ x = \frac{{4 \pm \sqrt{36}}}{{2}} ] [ x = \frac{{4 \pm 6}}{{2}} ]
This gives us two critical points: [ x_1 = 5 ] [ x_2 = 1 ]

Now, we'll use the second derivative test to determine the nature of these critical points. [ f''(x) = 6x  12 ]

Evaluate ( f''(5) ) and ( f''(1) ): [ f''(5) = 6(5)  12 = 30  12 = 18 ] [ f''(1) = 6(1)  12 = 6  12 = 18 ]

Since ( f''(5) > 0 ), ( x = 5 ) corresponds to a local minimum, and since ( f''(1) < 0 ), ( x = 1 ) corresponds to a local maximum.
Therefore, ( f(x) ) has a local minimum at ( x = 5 ) and a local maximum at ( x = 1 ).
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