"Find the absolute maximum and absolute minimum values..." solved and explained

ediculeN

Answered question

2021-05-21

Find the absolute maximum and absolute minimum values of f over the interval.

f (x) = (\frac{4}{x}) + \ln (x^{2}), 1 \leq x \leq 4

Answer & Explanation

Bentley Leach

Skilled2021-05-22Added 109 answers

Step 1
Given,
$f (x) = \frac{4}{x} + \ln (x^{2}), 1 \leq x \leq 4$
Step 2
The absolute maximum or absolute minimum values of a function exist at a point where its first derivative is zero or at the end points of the given interval.
Now differentiating given function with respect to x, we get
$f^{'} (x) = - \frac{4}{x^{2}} + \frac{1}{x^{2}} \cdot 2 x$
$\Rightarrow f^{'} (x) = - \frac{4}{x^{2}} + \frac{2}{x}$
Now $f^{'} (x) = 0$
$\Rightarrow - \frac{4}{x^{2}} + \frac{2}{x} = 0$
$\Rightarrow - 4 + 2 x = 0$
$\Rightarrow 2 x = 4$
$\Rightarrow x = 2$
Step 3
Now computing the value of given function at $x = 1$ , $x = 4$ (end points) and at $x = 2$ (point at which first derivative is zero).
$f (1) = \frac{4}{1} + \ln (1^{2}) = 4$
$f (2) = \frac{4}{2} + \ln (2^{2}) = 3.386294$
$f (4) = \frac{4}{4} + \ln (4^{2}) = 3.772588$
Step 4
Therefore absolute minimum value is 3.386294 and absolute maximum value is 4.

Jeffrey Jordon

Expert2021-11-19Added 2605 answers

Answer is given below (on video)

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