The absolute minimum value of f(x)=(x33)−(5x22)−6x on [-1,3] is?

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High School
Algebra
Algebra I
Piecewise-Defined Functions

Monique Slaughter

Answered question

2021-12-13

The absolute minimum value of

f (x) = (\frac{x^{3}}{3}) - (5 \frac{x^{2}}{2}) - 6 x

on [-1,3] is?

Answer & Explanation

chumants6g

Beginner2021-12-14Added 33 answers

Given,
$f (x) = \frac{x^{3}}{3} - \frac{5 x^{2}}{2} - 6 x, [- 1, 3]$
Differentiating with respect to x, we get
$f^{'} (x) = \frac{1}{3} (3 x^{2}) - \frac{5}{2} (2 x) - 6 (1)$
$= x^{2} - 5 x - 6$
Now to find critical values, solve the first derivative by equating it to zero. That is,
f'(x)=0
$\Rightarrow x^{2} - 5 x - 6 = 0$
$\Rightarrow x^{2} - 6 x + x - 6 = 0$
$\Rightarrow x (x - 6) + (x - 6) = 0$
$\Rightarrow (x - 6) (x = 1) = 0$
$\Rightarrow x = 6 or x = - 1$
But $6 n o t \in$ [-1,3], therefore critical value is x=-1.
Step 2
Now we find the value of the function at the critical value and at the end points of the given interval.
When x=-1,
$f (- 1) = \frac{{(- 1)}^{3}}{3} - \frac{5 {(- 1)}^{2}}{2} - 6 (- 1)$
$= - \frac{1}{3} - \frac{5}{2} + 6$
$= \frac{- 2 - 15 + 36}{6}$
$= \frac{19}{6}$
When x=3,
$f (3) = \frac{{(3)}^{3}}{3} - \frac{5 {(3)}^{2}}{2} - 6 (3)$
$= 9 - \frac{45}{2} - 18$
$= \frac{18 - 45 - 36}{2}$
$= - \frac{63}{2}$
Hence absolute minimum value is $- \frac{63}{2}$ which occurs at x=3.

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