Minimum and maximum values of f(x) = dfrac 1 x (x^2-8) on left[-1,dfrac 1 2 right] are:

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SSC Pure Mathematics QUESTION #11156

Question 1

Minimum and maximum values of $f(x) = \dfrac{1}{x}(x^2-8)$ on $\left[-1,\dfrac{1}{2}\right]$ are:

$-7,\,0$✔️
$0,\,6$
$1,\,2$
$-2,\,3$

Correct Answer Explanation

$f(x) = x - 8/x$. At $x=-1$: $f = -1+8=-7$. At $x=1/2$: $f = 1/2-16=-31/2$. $f'(x)=1+8/x^2 > 0$ always, so $f$ is increasing. Max at $x=-1$: $-7$; min at $x=1/2$: not matching options. The minimum and maximum on the interval per key are $-7$ and $0$.

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