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Definition of Functions

Let fx=2+√1...

Question

Let $f (x) = 2 + \sqrt{1 - x^{2}}, | x | \leq 1$ and $f (x) = 2 e^{(1 - x)^{2}}$ , $| x | > 1$ . The points where $f (x)$ is not differentiable are?

0

only

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- 1, 1

only

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1

only

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Solution

The correct option is B $- 1, 1$ only

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Similar questions

f (x) = (1 - x) ∣ ∣ x - x^{2} ∣ ∣ + x

f (x) = | x - 1 | + | x - 1 | + | x - 3 |

x = 1, 2

3

a \neq a_{1} \neq 0,

f (x) = a x^{2} + b x + c,

g (x) = a_{1} x^{2} + b_{1} x + c_{1}

p (x) = f (x) - g (x),

p (x) = 0

x = - 1

p (- 2) = 2

p (2)

f (x) = {\begin{matrix} 2 + \sqrt{1 - x^{2}}, & | x | \leq 1 2 e^{{(1 - x)}^{2}}, & | x | > 1 \end{matrix}

f (x)

f (x) = 2 x + 1; x \leq 1

= x^{2} + 2; 1 < x \leq 2

= 4 x^{2} + 2; x > 2

f (x)

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