# Consider the function f(x)=2x^{3}-6x^{2}-18x+9 on the interval [-2,4]. What is

Consider the function $$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}-{6}{x}^{{{2}}}-{18}{x}+{9}$$ on the interval [-2,4].
What is the absolute minimum of f(x) on [-2,4]?
What is the absolute maximum of f(x) on [-2,4]?

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Theodore Schwartz
Step 1
Consider the function
$$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}-{6}{x}^{{{2}}}-{18}{x}+{9}$$, on [-2,4]
Step 2
The given function is a polynomial so it is continuous everywhere therefore is continuous on [-2,4]
Critical points: Critical points of the function are the points where the derivative of the function either zero or does not exist.
Compute the derivative of the function
$$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}-{6}{x}^{{{2}}}-{18}{x}+{9}$$
$$\displaystyle{f}'{\left({x}\right)}={6}{x}^{{{2}}}-{12}{x}-{18}$$
To find the critical points put f'(x)=0.
f'(x)=0
$$\displaystyle{6}{x}^{{{2}}}-{12}{x}-{18}={0}$$
$$\displaystyle{6}{\left({x}^{{{2}}}-{2}{x}-{3}\right)}={0}$$
$$\displaystyle{x}^{{{2}}}-{2}{x}-{3}={0}$$
(x-3)(x+1)=0
x=-1,3
There are two critical points x=-1 and x=3 and both lies in the interval [-2,4].
Step 3
Evaluate the function value at the critical points and end points.
$$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}-{6}{x}^{{{2}}}-{18}{x}+{9}$$
at x=-2
$$\displaystyle{f{{\left(-{2}\right)}}}={2}{\left({2}\right)}^{{{3}}}-{6}{\left(-{2}\right)}^{{{2}}}-{18}{\left(-{2}\right)}+{9}$$
f(-2)=5
at x=-1
$$\displaystyle{f{{\left(-{1}\right)}}}={2}{\left(-{1}\right)}^{{{3}}}-{6}{\left(-{1}\right)}^{{{2}}}-{18}{\left(-{1}\right)}+{9}$$
f(-1)=19
at x=3
$$\displaystyle{f{{\left({3}\right)}}}={2}{\left({3}\right)}^{{{3}}}-{6}{\left({3}\right)}^{{{2}}}-{18}{\left({3}\right)}+{9}$$
f(3)=-45
at x=4
$$\displaystyle{f{{\left({4}\right)}}}={2}{\left({4}\right)}^{{{3}}}-{6}{\left({4}\right)}^{{{2}}}-{18}{\left({4}\right)}+{9}$$
f(4)=-31
Absolute maximum value and minimum value is the largest and the smallest function value respectively.
Absolute Maximum value=19 at x=-1
Absolute Minimum value=-45 at x=-3
Step 4
Absolute Maximum value= 19
Absolute Minimum value= -45