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### Answers (1)

Answer:

Absolute Maximum = 25 at $x=0,$

Absolute Minimum = -39 at $x=-2$

Hint:

Find the critical points and check the value of $f(x)$

Given:

$f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25 \text { in }[0,3]$

Explanation:

$f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48$

\begin{aligned} &f^{\prime}(x)=0 \\ &12 x^{3}-24 x^{2}+24 x-42=0 \\ &x^{3}-2 x^{2}+2 x-4=0 \\ &x^{2}(x-2)+2(x-2)=0 \\ &\left(x^{2}+2\right)(x-2)=0 \\ &x=2 \end{aligned}

Now, we check the value of $f(x)$ at $x=0, 2, 3$

\begin{aligned} &f(0)=3 \times 0-2 \times 8+12 \times 0-48 \times 0+25=25 \\ &f(2)=3 \times(2)^{4}-8 \times(2)^{3}+12 \times(2)^{2}-48 \times 2+25 \\ &=48-64+48-96+25 \\ &=-39 \end{aligned}

\begin{aligned} &f(3)=3 \times(3)^{4}-8 \times(3)^{3}+12 \times(3)^{2}-48 \times 3+25 \\ &=243-216+108-144+25 \\ &=16 \end{aligned}

Hence,

Absolute maximum = 25 at $x=0,$

Absolute minimum = -39 at $x=-2$

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