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If $f(x)$ is defined as follows, what is the minimum value of $f(x)$ for $x \in (0, 2]$ ?

$$f(x) = \begin{cases}  \frac{25}{8x} &\text{ when } x \leq \frac{3}{2} \\ x+ \frac{1}{x} &\text { otherwise}\end{cases}$$

  1. $2$
  2. $2 \frac{1}{12}$
  3. $2\frac{1}{6}$
  4. $2\frac{1}{2}$
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5 Answers

Best answer
42 votes
42 votes

$$f'(x) = \begin{cases}  \frac{-25x^{-2}}{8} \text{;  } x \leq \frac{3}{2} \text{,             decreasing in this interval}\\ 1- \frac{1}{x^2} \text {;  otherwise,     increasing in this interval}\end{cases}$$

$$f''(x) > 0, x \in \left(0, \frac{3}{2}\right]\text{;  }  \text{ $f(x)$ will be  minimum at $x = \frac{3}{2}$}\\ f''(x) < 0 \text {, otherwise; $\ \ \ $  $f(x)$ will  be maximum at $x = 2$}$$

$f(x)$ is decreasing in $(0, \frac{3}{2}]$ and in $\left(\frac{3}{2},  2\right]$ it is increasing.
So, minimum value of $f(x) $ must be at $x = \frac{3}{2}$
$f(\frac{3}{2}) = \frac{25}{12} = 2 \frac{1}{12}$

selected by
16 votes
16 votes
Answer: B

at x = $\frac{3}{2},$

f(x) =$2\left(\frac{1}{12}\right).$
edited by
9 votes
9 votes

Answer is B. The solution is as follows-

Solution

5 votes
5 votes

$f(x)=\frac{25}{8x}$   when x<=$\frac{3}{2}$

        $=x+\frac{1}{x}$   otherwise

Now if we check continuity at x=3/2,

Left hand limit ,

$\lim_{x->\frac{3}{2}^{-}}f(x)$

=$\lim_{x->\frac{3}{2}^{-}}\frac{25}{8x}$

=$\frac{25}{8}*\frac{2}{3}$

=$\frac{25}{12}$

Right hand limit,

$\lim_{x->\frac{3}{2}^{+}}f(x)$

=$\lim_{x->\frac{3}{2}^{+}}x+\frac{1}{x}$

=$\frac{3}{2}+\frac{2}{3}$

=$\frac{13}{6}$

so, The function is not continuous at $x=\frac{3}{2}$ so it is not differentiable in the range $(0,2]$ .

---------------------------------------------------------------------------------------------------------------------------------------------------

Now finding the minimum value using differentiation in the range is not correct as it is not differentiable in the range.

one important result want to point out

a function can have local maxima or minima at not critical point.For strictly monotonically increasing or decreasing function has the maximum or minimum value lies in the endpoints of the interval.

If we look carefully 

$f(x)=\frac{3.125}{x}$   when x<=$\frac{3}{2}$

        $=x+\frac{1}{x}$   otherwise

then f(x) is decreasing in the range $(0,\frac{3}{2}]$  and increasing in the range $(\frac{3}{2},2]$.

As when x is in the range  $(0,\frac{3}{2}]$ then ,as x>0 and increases, then function $\frac{3.185}{x}$ so , f(x) decreases till x=3/2 .

Now when x is in the range $(\frac{3}{2},2]$ then as x>3/2 and increases ,then function f(x)=$x+\frac{1}{x}$ is increases.

so the function has minimum point in x=3/2 as till x=3/2 the function decreases then increases and maximum point at x=2 in $(0,2]$ .

so , minimum value 

f(3/2)=$\frac{25}{12}$

the maximum value 

f(2)=$2+\frac{1}{2}=2.5$

So correct answer is (B).

Answer:

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