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+8 votes

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. $2\left(\frac{1}{12}\right)$
  3. $2\left(\frac{1}{6}\right)$
  4. $2\left(\frac{1}{2}\right)$
asked in Calculus by Boss (19.1k points)
edited by | 1.6k views

3 Answers

+13 votes
Best answer
Answer: B

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

f(x) =$2\left(\frac{1}{12}\right).$
answered by Boss (34.1k points)
edited by
how f(3/2) = 2/12 ????
shouldn't be 25/12 ??
$\frac{25}{12}$ = 2$\frac{1}{12}$

can we say that function is decreasing in the interval x(,3/2] and then increasing after 3/2, so the possible minima is at 3/2???

if we use the property of AM(arithmetic mean ) and GM(geometric mean) i.e.Am>=gm

then AM of X and 1/X=( (X+1/X)/2) and  GM of X and 1/X=1 so by AM$\geq$GM,we get

X+1/X$\geq$2,so ans should be minimum value is 2,correct me If I am wrong anywhere
+21 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 $x = \frac{3}{2}$
$f(\frac{3}{2}) = \frac{25}{12} = 2 \frac{1}{12}$

answered by Veteran (59.5k points)
edited by
f"(x) > 0 for x belongs to (x   3/2]

How did you  solved f'(x)=0 ?
How are you saying -25/8x^2 is decreasing for (0, 3/2]. It is minus sign, and x^2 is in denominator, so greater the value of x, lesser the value of 25/8x^2; since minus is before, which means the values INCREASE as -1 > -2(example)!!
please clear it more

'(x) = -25/8x^2 

f ''(x) = 50/8x^3>0 so f(x) is minimum b/w  (x   3/2] .i agree.

'(x)  =1 - 1/x^2  

but f ''(x) = 2/x^3>0 blw  (3/2 2] .how can say it is maximum.plz explain.

@Digvijay $f(x)$ is maximum at $x=2$? This need not be true based on the given arguments. It can be maximum at the start too.
There is also a point

this function has maximum value at $\propto$

$f''(x)=+\frac{50}{8x^{3}}$  when $x\leq \frac{3}{2}$


$f''(x)=\frac{2}{x^{3}}$ otherwise

So, both cases we get minimum value

Now, putting value $x=\frac{3}{2}$ in $f(x)$ we get $2\frac{1}{12}$...........................$i$

and putting 2 which satisfies function $f(x)=x+\frac{1}{x}$ we get $2\frac{1}{2}$................$ii$

Now $(i)$ satisfies minimum among them
Best explanation
–2 votes
2(1/12) NOT 2
answered by Active (3.3k points)

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