"At where f(x)=|(x2)−9| is differentiable?" solved and explained

Kymani Patrick

Answered question

2022-03-18

At where

f (x) = | (x^{2}) - 9 |

is differentiable?

Roland Ramsey

Beginner2022-03-19Added 4 answers

Since

x^{2} - 9

is a polynom it is differentiable everywhere. Then, in first approximation

| x^{2} - 9 |

is differentiable everywhere except when

x^{2} - 9 = 0

,
which is

x = \pm 3

Now we must calculate derivates at the points

- 3^{+}, - 3^{-}

and

3^{+}, 3^{-}

, and see if the function derivate is continuous.
If

x^{2} - 9 > 0, f (x) = x^{2} - 9

, and f'(x)=2x
If

x^{2} - 9 < 0, f (x) = - x^{2} + 9

, and f'(x)=-2x
When x=3:

f^{'} (3^{-}) = - 2 \cdot 3 = - 6

f^{'} (3^{+}) = 2 \cdot 3 = 6

The values are different, therefore f(x) is not differentiable in x=3.
The same conclusion could be obtained when x=-3.

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