Question

Find the absolute maximum and minimum values of a function $f$ given by $f\left(x\right)=2x^3-15x^2+36x+1$ on the interval $[1,5]$.

Answer 1

$f\left(x\right)=2x^3-15x^2+36x+1$
Differentiating w.r.t. $x,$
$f^{\prime }\left(x\right)=6x^2-30x+36$
$f^{\prime }\left(x\right)=0$
$6x^2-30x+36=0$
$\Rightarrow 6(x^2-5x+6)=0$
$\Rightarrow 6(x-2)(x-3)=0$
$\Rightarrow x=2,3$
Critical points are $x=2,3$
Finding the function's value at $x=2,3$ and endpoints $x=1,5$
$f\left(x\right)=2x^3-15x^2+36x+1$
$f\left(1\right)=2\left(1^3\right)-15\left(1^2\right)+36\left(1\right)+1=24$
$f\left(2\right)=2\left(2^3\right)-15\left(2^2\right)+36\left(2\right)+1=29$
$f\left(3\right)=2\left(3^3\right)-15\left(3^2\right)+36\left(3\right)+1=28$
$f\left(5\right)=2\left(5^3\right)-15\left(5^2\right)+36\left(5\right)+1=56$
Hence, the absolute maximum value is $56$ at $x=5$ and absolute minimum value is $24$ at $x=1$.