Calculus

Calculus Level 4

The function f f is defined by y = f ( x ) y = f(x) , where x = 2 t − ∣ t ∣ x =2t - |t| and y = t 2 + t ∣ t ∣ y = t^2 + t | t| for t ∈ R t\in \mathbb R .

Is f ( x ) f(x) continuous and/or differentiable at x = 0 x=0 ?

Neither continuous nor differentiable at x = 0 x=0 Discontinuous at x = 0 x=0 Non-differentiable at x = 0 x=0 Continuous and differentiable at x = 0 x=0

Hemantika Sengar by Hemantika Sengar , 18, India

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1 solution

Chew-Seong Cheong
Jun 21, 2016

{ x = 2 t − ∣ t ∣ = { 3 t if t < 0 t if t ≥ 0 y = t 2 − t ∣ t ∣ = { 0 if t < 0 2 t 2 if t ≥ 0 \begin{cases} x = 2t - |t| & = \begin{cases} 3t & \ \ \text{if } t < 0 \\ t & \ \ \text{if } t \ge 0 \end{cases} \\ y = t^2 - t|t| & = \begin{cases} 0 & \text{if } t < 0 \\ 2t^2 & \text{if } t \ge 0 \end{cases} \end{cases}

⟹ f ( x ) = y = { 0 if x < 0 2 x 2 if x ≥ 0 \implies f(x) = y = \begin{cases} 0 & \text{if } x < 0 \\ 2x^2 & \text{if } x \ge 0 \end{cases}

{ f ( 0 − ) = 0 f ( 0 + ) = 0 ⟹ f ( 0 − ) = f ( 0 + ) Continuous at x = 0 \begin{cases} f(0^-) = 0 \\ f(0^+) = 0 \end{cases} \implies f(0^-) = f(0^+) \quad \text{Continuous at }x=0

{ f ′ ( 0 − ) = 0 f ′ ( 0 + ) = 0 ⟹ f ′ ( 0 − ) = f ′ ( 0 + ) Differentiable at x = 0 \begin{cases} f'(0^-) = 0 \\ f'(0^+) = 0 \end{cases} \implies f'(0^-) = f'(0^+) \quad \text{Differentiable at }x=0

Therefore, f ( x ) f(x) is continuous and differentiable at x = 0 . \boxed{\text{continuous and differentiable at }x=0}.

4 Helpful 0 Interesting 1 Brilliant 0 Confused

Is it langrangian mean value therem ? that differentiability and continuty at x = 0

A Former Brilliant Member - 4 years, 9 months ago

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No, I did not use mean value theorem. I just check if the differentiation results from both sides. My solution may not be adequate.

Chew-Seong Cheong - 4 years, 9 months ago

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