Describe the x-values at which the function is differentiable. (enter your answer using interval notation.) y = |x² - 25|

Solution:

Given, y = |x² - 25|

{x² - 25 if x ϵ (-∞, -5) U (5, ∞)

y = {0 if x = 5/-5

{25 - x² if x ϵ (-5, 5)

Case (i): Differentiability at x = 5

L.H.D = lim_x→5⁻ [(f(x) - f(5)) / (x - 5)] 

= lim_x→5⁻ [((25 - x²) - 0) / (x - 5)] 

= lim_x→5⁻ [((5 - x)(5 + x)) / (x - 5)]

= lim_x→5⁻ -(5 + x)

= -(5 + 5)

= -10

R.H.D = lim_x→5⁺ [(f(x) - f(5)) / (x - 5)] 

= lim_x→5⁺ [((x² - 25) - 0) / (x - 5)] 

= lim_x→5⁺ [((x + 5)(x - 5)) / (x - 5)]

= lim_x→5⁺ (x + 5)

= (5 + 5)

= 10

⇒ L.H.D ≠ R.H.D

∴ f(x) is not differentiable at x = 5.

Case (ii) : Differentiability at x = -5

L.H.D = lim_x→5^- [(f(x) - f(-5)) / (x - (-5))] 

= lim_x→5^- [((x² - 25) - 0) / (x + 5)] 

= lim_x→5^- [((x + 5)(x - 5)) / (x + 5)]

= lim_x→5^- (x - 5)

= -5 - 5 

= -10

R.H.D = lim_x→5⁺ [(f(x) - f(-5)) / (x - (-5))] 

= lim_x→5⁺ [((25 - x²) - 0) / (x + 5)] 

= lim_x→5⁺ [((5 - x)(5 + x)) / (x + 5)]

= 5 - (-5)

= 10

⇒ L.H.D ≠ R.H.D

∴ f(x) is differential in (-∞, ∞) - {5, -5}

Note: 

⃒f(x)⃒ is not differentiable for all those values of ‘x’ at which f(x) = 0.

Given function |x² - 25|, f(x) = x² - 25 is zero for x = 5 and x = -5

∴ f(x) is not differentiable at x = 5 and x = -5.

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Describe the x-values at which the function is differentiable. (enter your answer using interval notation.) y = |x² - 25|

Summary:

The x-values at which the function y = |x² - 25| is differentiable is, in (-∞, ∞) - {5, -5}.