Get the solution to "find the absolute maximum and absolute minimum values..."

High School
Algebra
Algebra I
Piecewise-Defined Functions

Brittney Lord

Answered question

2020-10-26

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

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

Answer & Explanation

Gennenzip

Skilled2020-10-27Added 96 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

Expert2021-11-11Added 2605 answers

Answer is given below (on video)

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