Question

Find the intervals in which the function $f$ given by $f (x) = 2 x^{2} - 3 x$ is strictly decreasing.

Accepted Answer

Note: Let $f (x) = 2 x^{2} - 3 x$ be continuous on $[a, b]$ and differentiable on the open interval $(a, b)$ . Then

$(1)$ $f$ is strictly increasing in $[a, b] if f^{'} (x) > 0$ for each $x \in (a, b)$

$(2)$ $f$ is strictly decreasing in $[a, b] if f^{'} (x) < 0$ for each $x \in (a, b)$

$f (x) = 2 x^{2} - 3 x$

Calculating $f' (x)$

$f^{'} (x) = 4 x - 3$

Putting $f^{'} (x) = 0$

$4 x - 3 = 0$

$4 x = 3$

$x = \frac{3}{4}$

Drawing point on number line

Point $x = \frac{3}{4}$ divide the real line into $2$ disjoint intervals

$(- \infty, \frac{3}{4}) and (\frac{3}{4}, \infty)$

Value of $x$ Intervals Sign of $f^{'} (x) = 4 x - 3 = 4 (x - \frac{3}{4})$ Nature of function $f$

$x < \frac{3}{4}$ $(- \infty, \frac{3}{4})$ $(-) < 0$ $[$ for $x = \frac{3}{6}$ , $f^{'} (\frac{3}{6}) = 4 (\frac{3}{6} - \frac{3}{4}) = - 1 < 0]$ $f$ is strictly decreasing

$x > \frac{3}{4}$ $(\frac{3}{4}, \infty)$ $(+) > 0$ $[$ for $x = 1, f^{'} (1) = 4 (1 - \frac{3}{4}) = 1 > 0]$ $f$ is strictly increasing

Hence $f$ is strictly decreasing in $(- \infty, \frac{3}{4})$ .

EXERCISE 6.2

Jharkhand Board Mathematics Solutions for EXERCISE 6.2

Questions from EXERCISE 6.2 with Hints & Solutions

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