Find the absolute maximum and minimum values of a function f give… - Vidyadip

Question

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

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Answer

Given,$ f(x) = 2x^3 = 15x^2 + 36x + 1$
$\therefore f'(x) = 6x^2- 30x + 36$
$=6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$
Now, that $f'(x) = 0$ gives $x = 2$ and $x = 3$
We shall now evaluate the value of $f$ at these points and at the ens points of the interval$ [1,5],$
i.e at $x = 1, 2, 3$ and $5$
$At x = 1, f(1) = 2(1^3) - 15(1^2) + 36(1) + 1 = 24$
$At x = 2, f(2) = 2(2^3) - 15(2^2) + 36(2) + 1 = 29$
$At x = 3, f(3) = 2(3^3) - 15(3^2) + 36(3) + 1 = 28$
$At x = 5, f(2) = 2(5^3) - 15(5^2) + 36(5) + 1 = 56$
Thus, We conclude that the absolute maximum value of on $[1, 5]$ is $56,$ occurring at $x = 5,$ and absolute minimum value of f on $[1, 5]$ is $24$ which occurs at $x = 1.$

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