Answer to Question #148338 in Calculus for Angelo

Question

Find the number in the interval (1/3, 2) such that the sum of the number and its reciprocal
is a maximum.

Answer 1

$$Solution$$

Let us consider $x$  be the number

And its reciprocal is $\frac1x$

Now the sum of the number and its reciprocal is $x+\frac1x$

Let $f(x)=x+\frac1x$  on $[\frac13, 2]$

Clearly f  is a rational function continues on $[\frac13, 2]$

Now we need to find the absolute maximum (large) and absolute minimum (small) values of $f(x)$ on $[\frac13, 2]$

A continuous function $f$ on a closed interval $[a,b]$ has both absolute maximum and minimum values. These are obtained by using following procedure:

(i) First all critical numbers of the function are found on$[a,b]$ 

(ii) Then $f$  is evaluated at the critical numbers and the end points $a$ and $b$

The largest and the smallest values among the values of the function computed above gives the absolute maximum and absolute minimum values of the function.

Differentiating $f(x)$ the expression for  with respect to $x$

$$f'(x)=\frac{d}{dx}\begin{bmatrix} x+\frac{1}{x} \end{bmatrix}\\ \frac{d(x)}{dx}+\frac{d(\frac{1}{x})}{dx}\ since\begin{pmatrix} \frac{d(x^n)}{dx}=nx^{n-1} \end{pmatrix}\\ =1-\frac{1}{x^2}$$

Since $f'(x)$ is defined for all $x$ in the given $[\frac{1}{3},2]$

The critical numbers are obtained by equating the first derivative to 0.

Thus 

$f'(x)=0\\ \implies 1-\frac{1}{x^2}=0\\ \implies \frac{1}{x^2}=1$

Out of these two critical numbers $x=-1$ does not belong to the given $[\frac13,2]$. Hence $x=-1$ is discarded.

Therefore the lone critical number for the given function in the given

$[\frac13,2]$ is $x=1$ .

Next the value of the function at the critical number and the end points is computed. These are

$f(\frac{1}{3})=\frac{1}{3}+3=\frac{10}{3}\\ f(1)=1+1=2\\ f(2)=1+\frac{1}{2}=\frac{5}{5}\\$

Therefore  $f$ attains the larger value at $x=\frac13$  in the given$[\frac13, 2]$.