In the following functions, is any real number .
Let
Find the total number of points at which is not differentiable.
Notation : denotes the absolute value function .
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h ( x ) = { max { f ( x ) , g ( x ) } min { f ( x ) , g ( x ) } = max { 1 − x , x 2 + 1 } = min { 1 + x , x 2 + 1 } if x ≤ 0 if x > 0
Equating f ( x ) = g ( x ) for x ≤ 0 and x > 0 ,
x ≤ 0 : 1 − x x 2 + x x ( x + 1 ) = x 2 + 1 = 0 = 0
⟹ f ( − 1 ) = g ( − 1 ) = 2 , f ( 0 ) = g ( 0 ) = 1 , when x < − 1 , g ( x ) > f ( x ) and when − 1 < x < 0 , f ( x ) > g ( x ) .
Similarly,
x > 0 : 1 + x x 2 − x x ( x − 1 ) = x 2 + 1 = 0 = 0
⟹ f ( 0 ) = g ( 0 ) = 1 , f ( 1 ) = g ( 1 ) = 2 , when 0 < x < 1 , g ( x ) < f ( x ) and when x > 1 , f ( x ) < g ( x ) .
Therefore, h ( x ) = ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ x 2 + 1 1 − x x 2 + 1 1 + x if x < − 1 if − 1 < x < 0 if 0 < x < 1 if x > 1
There are three critical points at x = − 1 , 0 , 1 .
Since ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x = − 1 x = 0 x = 1 ⟹ { f ′ ( − 1 ) = − 1 g ′ ( − 1 ) = − 2 ⟹ { f ′ ( 0 ) = − 1 g ′ ( 0 ) = 0 ⟹ { f ′ ( 1 ) = 1 g ′ ( 1 ) = 2 ⟹ f ′ ( − 1 ) = g ′ ( − 1 ) ⟹ f ′ ( 0 ) = g ′ ( 0 ) ⟹ f ′ ( 1 ) = g ′ ( 1 ) Not differentiable Not differentiable Not differentiable
Therefore, there are 3 points at which h ( x ) is not differentiable.