If $f(x)$ is defined as follows, what is the minimum value of $f(x)$ for $x \in (0, 2]$ ?
$$f(x) = \begin{cases} \frac{25}{8x} \text{ when } x \leq \frac{3}{2} \\ x+ \frac{1}{x} \text { otherwise}\end{cases}$$
$$f'(x) = \begin{cases} \frac{-25x^{-2}}{8} \text{; } x \leq \frac{3}{2} \text{, decreasing in this interval}\\ 1- \frac{1}{x^2} \text {; otherwise, increasing in this interval}\end{cases}$$
$$f''(x) > 0, x \in \left(0, \frac{3}{2}\right]\text{; } \text{ $f(x)$ will be minimum at $x = \frac{3}{2}$}\\ f''(x) < 0 \text {, otherwise; $\ \ \ $ $f(x)$ will be maximum at $x = 2$}$$ $f(x)$ 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 $x = \frac{3}{2}$ $f(\frac{3}{2}) = \frac{25}{12} = 2 \frac{1}{12}$
f '(x) = -25/8x^2
f ''(x) = 50/8x^3>0 so f(x) is minimum b/w (x 3/2] .i agree.
f '(x) =1 - 1/x^2
but f ''(x) = 2/x^3>0 blw (3/2 2] .how can say it is maximum.plz explain.
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???