# Find the absolute maximum and absolute minimum values of function on the given interval. f(x)=5+54x-2x^3, [0,4]

Question
Piecewise-Defined Functions
Find the absolute maximum and absolute minimum values of function on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}},{\left[{0},{4}\right]}$$

2021-03-07
Step 1
Given:
Given that function $$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}}$$
The given interval is $$\displaystyle{\left[{0},{4}\right]}$$
To find:
we have to find the absolute maximum and absolute minimum
Step 2
Here $$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}}$$ ...(1) and the interval is $$\displaystyle{\left[{0},{4}\right]}$$
Differentiate (1) w.r.t.x, we get
$$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({5}+{54}{x}-{2}{x}^{{3}}\right)}$$
$$\displaystyle{f}'{\left({x}\right)}={54}-{6}{x}^{{2}}$$
set f'(x)=0
$$\displaystyle\Rightarrow{54}-{6}{x}^{{2}}={0}$$
$$\displaystyle\Rightarrow{6}{x}^{{2}}={54}$$
$$\displaystyle\Rightarrow{x}^{{2}}={9}$$
$$\displaystyle\Rightarrow{x}={3}$$ and x=-3
but $$\displaystyle{x}=-{3}\notin{\left[{0},{4}\right]}$$
and it also contain the critical point x =0
Step 3
Hence, the critical points is x = 3 and x = 0
Now, $$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}}$$
For x = 3
$$\displaystyle{f{{\left({3}\right)}}}={5}+{54}{\left({3}\right)}-{2}{\left({3}\right)}^{{3}}={5}+{162}-{54}$$
=113
Thus, the absolute value is maximum at x = 3
Now, $$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}}$$
For x = 0
f(0)=5
Thus, the absolute value is minimum at x =0

### Relevant Questions

Find the absolute maximum and absolute minimum values of f on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{{3}}},{\left[{0},{4}\right]}$$
Find the absolute maximum and absolute minimum values of f on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={5}+{54}{x}-{2}{x}^{{3}},{x}\in{\left[{0},{4}\right]}$$
Find the absolute maximum and absolute minimum values of the function:
$$\displaystyle{f{{\left({x}\right)}}}={5}{x}^{{{7}}}−{7}{x}^{{{5}}}−{7}$$ on the interval $$\displaystyle{\left[−{3},{4}\right]}$$
Find the absolute maximum and absolute minimum values of f on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={x}+\frac{{4}}{{x}},{\left[{0.2},{8}\right]}$$
absolute minimum value-?
absolute maximum value-?
Find the absolute maximum and absolute minimum value of f on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={9}+{81}{x}-{3}{x}^{{3}},{\left[{0},{4}\right]}$$
Find the absolute maximum and absolute minimum values of f on the given interval.
$$\displaystyle{f{{\left({x}\right)}}}={4}{x}^{{3}}-{6}{x}^{{2}}-{24}{x}+{9},{\left[-{2},{3}\right]}$$
absolute minimum value-?
absolute maximum value-?
Find the absolute maximum and absolute minimum values of f on the given interval.
$$\displaystyle{f{{\left({t}\right)}}}={5}{t}+{5}{\cot{{\left(\frac{{t}}{{2}}\right)}}},{\left[\frac{\pi}{{4}},{7}\frac{\pi}{{4}}\right]}$$
absolute minimum value-?
absolute maximum value-?
$$\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({x}^{{{2}}}+{5}{x}+{13}\right)}}},{\left[−{3},{1}\right]}$$
$$\displaystyleƒ{\left({x}\right)}={\left(\frac{{4}}{{x}}\right)}+{\ln{{\left({x}^{{2}}\right)}}},{1}\le{x}\le{4}$$
Consider the function $$\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}−{72}{x}^{{{2}}}+{9},-{5}\leq{x}\leq{13}$$.