Question

Find the absolute maximum and absolute 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

$f\left(x\right)=2x^3-15x^2+36x+1$

$f^{\prime }\left(x\right)=6x^2-30x+36$

Putting $f^{\prime }\left(x\right)=0$

$6x^2-30x+36=0$

$x^2-5x+6=0$

$(x-2)(x-3)=0$

$x=2$ or $3$

We are given interval $[1,5]$

Hence, calculating f(x) at 2, 3, 1, 5,

$f\left(2\right)=29$

$f\left(3\right)=28$

$f\left(1\right)=24$

$f\left(5\right)=56$

Hence, absolute maximum value is $56$ at $x=5$.

Absolute minimum value is $24$ at $x=1$.