Question

Let $f\left(x\right)=3x^3-7x^2+5x+6.$ The maximum value of $f\left(x\right)$ over the interval $[0,2]$ is (up to $1$ decimal place).

• A. 12

Answer 1

The correct option is A 12
$f\left(x\right)=3x^2-7x^2+5x+6$ in $[0,2]$
$f^{\prime }\left(x\right)=9x^2-14x+5$
and $f^{\prime \prime }\left(x\right)=18x-14$
Critical Points are $f^{\prime }\left(x\right)=0\Rightarrow x=1$ and $dfrac59$
$\because f^{\prime \prime }\left(1\right)=+ve$ so $x=1$ in point of minimum
$f^{\prime \prime }\left(\frac{5}{9}\right)=-ve$ so $x=\frac{5}{9}$ is point of maximum and value
$=f\left(\frac{5}{9}\right)=3{\left[\frac{5}{9}\right]}^3-7{\left[\frac{5}{9}\right]}^2+5\left[\frac{5}{7}\right]+6$
$=7.1$
But maximum value can also occur at corner points.
So
$f\left(2\right)=3(2{)}^3-7(2{)}^2+5\left(2\right)+6=12$
So absolute maximum value $=12.$