GATE IT 2008 | Question: 31

Question

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

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

• A. $2$
• B. $2 \frac{1}{12}$
• C. $2\frac{1}{6}$
• D. $2\frac{1}{2}$

Comments

The given function is neither continuous nor differentiable at x=3/2. But still, the minima exists on 3/2. But I have a doubt. If the function is not continuous at 3.2, then how the minima at that point existing?

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Good Explanation 

https://gateoverflow.in/3341/gate-it-2008-question-31?show=362595#a362595

Answer 1

2(1/12) NOT 2

Comments

Why not 2 is the minimum value ?
Because if we take second one it will be y = x + 1/x  
>> 1- 1(x^2)
>> x = 1 (for min), x= -1 (for max)
>> y(1) = 1+1 = 2

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bcoz x=1 doesn’t belongs to domain (3/2,2], so you can’t use x + 1/x