The sum of the maximum and the minimum values of $3 x^4-2 x^3-6 x^2+6 x+4$, in $(0,2)$ is
The sum of the maximum and the minimum values of $3 x^4-2 x^3-6 x^2+6 x+4$, in $(0,2)$ is
- 28
- $\frac{167}{16}$
- $\frac{134}{15}$
- $\frac{87}{16}$
Solution
Let a function $f(x)=3 x^4-2 x^3-6 x^2+6 x+4$
So,
$
\begin{aligned}
f^{\prime}(x) & =12 x^3-6 x^2-12 x+6 \\
& =6(x-1)(x+1)(x-1 / 2
\end{aligned}
$
For maxima and minima
$
\begin{array}{rlrl}
f^{\prime}(x) & =0 \\
\Rightarrow \quad x & =1, \frac{1}{2}, \quad[\because-1 \notin(0,2] \\
& \text { Now, } \\
\text { and } & \quad f(1) & =5 \text { (minimum) } \\
& & f(1 / 2) & =\frac{87}{16} \text { (maximum) }
\end{array}
$
Now, $\quad f(1)=5$ (minimum)
and
$
f(1 / 2)=\frac{87}{16} \text { (maximum) }
$
So, $\quad f(1)+f\left(\frac{1}{2}\right)=\frac{167}{16}$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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