Check the solution to "Examine the continuity and differential possibilities of function..."

Falak Kinney

Answered question

2021-10-31

Examine the continuity and differential possibilities of function

f (x) = | x^{2} - 1 |

in x=1

Answer & Explanation

timbalemX

Skilled2021-11-01Added 108 answers

Given:

f (x) = | x^{2} - 1 |

To check for continuity at x=1, we find the limit to the left and right of x=1

lim_{x \to 1} | x^{2} - 1 |

x = 0.9

| {(0.9)}^{2} - 1 | = 0.19

x = 0.99

| {(0.99)}^{2} - 1 | = 0.0190

Left limit

lim_{x \to 1^{-}} | x^{2} - 1 | = 0

x=1.1

| {(1.1)}^{2} - 1 | = 0.21

x = 0.99

| {(0.99)}^{2} - 1 | = 0.2321

Right limit

lim_{x \to 1^{+}} | x^{2} - 1 | = 0

Both left and right limits are equal to 0
So the f(x) is continuous at x=1
To check fordifferetiability, we use derivative
Lets find the derivative of f(x).
We apply the derivative of absolute function and chain rule

f (x) = | x |

f^{'} (x) = \frac{x}{| x |}

f (x) = | x^{2} - 1 |

f^{'} (x) = \frac{x^{2} - 1}{| x^{2} - 1 |} \cdot (2 x)

f^{'} (x) = \frac{2 x (x^{2} - 1)}{| x^{2} - 1 |}

Now we plug in x=1 in the derivative

f^{'} (x) = \frac{2 x (x^{2} - 1)}{| x^{2} - 1 |}

f^{'} (1) = \frac{2 (1) (1^{2} - 1)}{| 1^{2} - 1 |} = \infty

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