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The maximum value of function $x^{3}-12 x^{2}+36 x+17$ in the interval $[1,10]$ is
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177
Let $f(x)=x^{3}-12 x^{2}+36 x+17$
$\therefore f^{\prime}(x)=3 x^{2}-24 x+36=0$
For maxima, put $f^{\prime}(x)=0$ $\Rightarrow \quad 3 x^{2}-24 x+36=0$
$\Rightarrow \quad(x-2)(x-6)=0$
$\Rightarrow \quad x=2,6$
Again, $f^{\prime \prime}(x)=6 x-24$ is negative at $x=2$ So that, $f(6)=17, f(2)=49$
At the end points, $f(1)=42, f(10)=177$ So that, $f(x)$ has its maximum value 177 .
$\therefore f^{\prime}(x)=3 x^{2}-24 x+36=0$
For maxima, put $f^{\prime}(x)=0$ $\Rightarrow \quad 3 x^{2}-24 x+36=0$
$\Rightarrow \quad(x-2)(x-6)=0$
$\Rightarrow \quad x=2,6$
Again, $f^{\prime \prime}(x)=6 x-24$ is negative at $x=2$ So that, $f(6)=17, f(2)=49$
At the end points, $f(1)=42, f(10)=177$ So that, $f(x)$ has its maximum value 177 .
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