SATHEE: Application of Derivatives 4 Question 61

Application of Derivatives 4 Question 61

####64. Let $f : R \to R$ be defined as $f (x) = | x | + | x^{2} - 1 |$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

(2012)

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Solution:

PLAN

(i) Local maximum and local minimum are those points at which $f^{'} (x) = 0$ , when defined for all real numbers

(ii) Local maximum and local minimum for piecewise functions are also been checked at sharp edges.

Description of Situation $y = | x | = \begin{aligned} x, if x \geq 0 \\ - x, if x < 0 \end{aligned}$

Also, $y = | x^{2} - 1 | = \begin{array}{ll} (x^{2} - 1), & if x \leq - 1 or x \geq 1 \\ (1 - x^{2}), & if - 1 \leq x \leq 1 \end{array}$

$y = | x | + | x^{2} - 1 | = \begin{array}{ll} - x + 1 - x^{2}, & if x \leq - 1 \\ - x + 1 - x^{2}, & if - 1 \leq x \leq 0 \\ x + 1 - x^{2}, & if 0 \leq x \leq 1 \\ x + x^{2} - 1, & if x \geq 1 \\ - x^{2} - x + 1, & if x \leq - 1 \\ - x^{2} - x + 1, & if - 1 \leq x \leq 0 \\ - x^{2} + x + 1, & if 0 \leq x \leq 1 \\ x^{2} + x - 1, & if x \geq 1 \end{array}$

which could be graphically shown as

Thus, $f (x)$ attains maximum at $x = \frac{1}{2}, \frac{- 1}{2}$ and $f (x)$ attains minimum at $x = - 1, 0, 1$ .

$\Rightarrow$ Total number of points $= 5$

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