# The absolute minimum value of f(x)=(x^{3}/3)-(5x^{2}/2)-6x on [-1,3] is?

The absolute minimum value of $f\left(x\right)=\left(\frac{{x}^{3}}{3}\right)-\left(5\frac{{x}^{2}}{2}\right)-6x$ on [-1,3] is?
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chumants6g

Given,
$f\left(x\right)=\frac{{x}^{3}}{3}-\frac{5{x}^{2}}{2}-6x,\left[-1,3\right]$
Differentiating with respect to x, we get
${f}^{\prime }\left(x\right)=\frac{1}{3}\left(3{x}^{2}\right)-\frac{5}{2}\left(2x\right)-6\left(1\right)$
$={x}^{2}-5x-6$
Now to find critical values, solve the first derivative by equating it to zero. That is,
f'(x)=0
$⇒{x}^{2}-5x-6=0$
$⇒{x}^{2}-6x+x-6=0$
$⇒x\left(x-6\right)+\left(x-6\right)=0$
$⇒\left(x-6\right)\left(x=1\right)=0$

But [-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,
$f\left(-1\right)=\frac{{\left(-1\right)}^{3}}{3}-\frac{5{\left(-1\right)}^{2}}{2}-6\left(-1\right)$
$=-\frac{1}{3}-\frac{5}{2}+6$
$=\frac{-2-15+36}{6}$
$=\frac{19}{6}$
When x=3,
$f\left(3\right)=\frac{{\left(3\right)}^{3}}{3}-\frac{5{\left(3\right)}^{2}}{2}-6\left(3\right)$
$=9-\frac{45}{2}-18$
$=\frac{18-45-36}{2}$
$=-\frac{63}{2}$
Hence absolute minimum value is $-\frac{63}{2}$ which occurs at x=3.