# Find both the maximum value and the minimum value of

Question:

Find both the maximum value and the minimum value of

$3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$ on the interval $[0,3]$

Solution:

Let $f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$

\begin{aligned} \therefore f^{\prime}(x) &=12 x^{3}-24 x^{2}+24 x-48 \\ &=12\left(x^{3}-2 x^{2}+2 x-4\right) \\ &=12\left[x^{2}(x-2)+2(x-2)\right] \\ &=12(x-2)\left(x^{2}+2\right) \end{aligned}

Now, $f^{\prime}(x)=0$ gives $x=2$ or $x^{2}+2=0$ for which there are no real roots.

Therefore, we consider only $x=2 \in[0,3]$.

Now, we evaluate the value of at critical point x = 2 and at the end points of the interval [0, 3].

\begin{aligned} f(2) &=3(16)-8(8)+12(4)-48(2)+25 \\ &=48-64+48-96+25 \\ &=-39 \\ f(0) &=3(0)-8(0)+12(0)-48(0)+25 \\ &=25 \\ f(3) &=3(81)-8(27)+12(9)-48(3)+25 \\ &=243-216+108-144+25=16 \end{aligned}

Hence, we can conclude that the absolute maximum value of on [0, 3] is 25 occurring at = 0 and the absolute minimum value of f at [0, 3] is − 39 occurring at x = 2.