Question

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

Answer 1

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