If f(x)=3x^3-9x^2-27x+15,then the maximum val | vedclass

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(B) Given the function $f(x) = 3x^3 - 9x^2 - 27x + 15$. To find the critical points,we calculate the first derivative: $f'(x) = 9x^2 - 18x - 27$. Sett

Vedclass · Accepted Answer

$30$

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(B) Given the function $f(x) = 3x^3 - 9x^2 - 27x + 15$. To find the critical points,we calculate the first derivative: $f'(x) = 9x^2 - 18x - 27$. Setting $f'(x) = 0$,we get $9(x^2 - 2x - 3) = 0$,which factors to $9(x - 3)(x + 1) = 0$. Thus,the critical points are $x = 3$ and $x = -1$. To determine the nature of these points,we find the second derivative: $f''(x) = 18x - 18$. At $x = 3$,$f''(3) = 18(3) - 18 = 36 > 0$,so $f(x)$ has a local minimum at $x = 3$. At $x = -1$,$f''(-1) = 18(-1) - 18 = -36 The maximum value is $f(-1) = 3(-1)^3 - 9(-1)^2 - 27(-1) + 15 = -3 - 9 + 27 + 15 = 30$.

If $f(x)=3x^3-9x^2-27x+15$,then the maximum value of $f(x)$ is $.....$

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