Get the solution to "Find the absolute extrema on the given interval...."

nuais6lfp

Answered question

2021-12-03

Find the absolute extrema on the given interval.

f (x) = x - \sqrt{3} {x}, [- 1, 4]

oces3y

Beginner2021-12-04Added 21 answers

Given function is

f (x) = x - \sqrt{3} {x}

and interval is [-1,4].
We need to find critical points of the function.
Differentiating the given function with respect to x,

f^{'} (x) = \frac{d}{d x} (x - \sqrt{3} {x})

= \frac{d}{d x} (x) - \frac{d}{d x} (x^{\frac{1}{3}})

= 1 - \frac{1}{3} x^{\frac{1}{3} - 1}

= 1 - \frac{1}{3} x^{- \frac{2}{3}}

Putting first derivative equals to zero.

1 - \frac{1}{3} x^{- \frac{2}{3}} = 0

\frac{1}{3} x^{- \frac{2}{3}} = 1

x^{- \frac{2}{3}} = 3

x = 3^{- \frac{3}{2}}

x = \frac{1}{3^{\frac{3}{2}}}

x = \frac{1}{3 \sqrt{3}}

Therefore, critical value of the function is

\frac{1}{3 \sqrt{3}}

.
Step 5

\frac{1}{3 \sqrt{3}}

lies in the interval [-1,4].
Now we will evaluate the function at the endpoints of the interval and critical value.

f (- 1) = 1 - \sqrt{3} {1}

=1-1
=0

f (\frac{1}{3 \sqrt{3}}) = 1 - \sqrt{3} {(3 \sqrt{3})}

= 1 - \sqrt{3} {5.19}

=1-2.27=-1.27
Step 6

f (4) = 4 - \sqrt{3} {4}

=4-1.58
=2.42
The absolute maximum value of the function is 2.42 at x=4.
The absolute minimum value of the function is -1.27 at

x = \frac{1}{3 \sqrt{3}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?