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Example · Example 8

Q.Find the local maximum and local minimum values of f(x)=x3−6x2+9x+15f(x) = x^3 - 6x^2 + 9x + 15 using the second derivative test.

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f(x)=x3−6x2+9x+15f(x)=x^3-6x^2+9x+15

f′(x)=3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3),f′′(x)=6x−12.f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3), \qquad f''(x)=6x-12.

Critical points: f′(x)=0  ⟹  x=1, 3f'(x)=0 \implies x=1,\ 3.

At x=1x=1: f′′(1)=6−12=−6<0  ⟹  f''(1)=6-12=-6<0 \implies local maximum. f(1)=1−6+9+15=19f(1)=1-6+9+15=19. …

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