Finding function continuity and derivatives

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SUMMARY

The function f defined by f(x) = { |x + 1|, x < 0; x^2 + 1, x ≥ 0 } is continuous on the intervals (-∞, -1), (-1, 0), and (0, +∞). It is differentiable on the intervals (-∞, -1) and (-1, 0) but not at x = -1 and x = 0 due to discontinuities in the derivative f'(x). The derivative f'(x) can be computed as f'(x) = { -1, x < -1; 2x, x > 0 } with f'(-1) and f'(0) being undefined.

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of limits and continuity
  • Familiarity with derivatives and differentiability
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of piecewise functions in calculus
  • Learn how to apply the limit definition of continuity
  • Explore the concept of differentiability and its implications
  • Practice finding derivatives of piecewise-defined functions
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Students studying calculus, mathematics educators, and anyone interested in understanding function continuity and derivatives in piecewise contexts.

ultima9999
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I'm not sure where to start with this question. If a limit was given, I could solve it but without it given, I am completely lost...

State on which intervals the function [tex]f[/tex] defined by [tex]f(x) = \left\{\begin{array}{cc}|x + 1|,&x < 0\\x^2 + 1,&x \geq 0\end{array}\right.[/tex] is:

i) continuous
ii) differentiable
Find the derivative [tex]f'(x)[/tex] at all points where the function is differentiable.
 
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It may be easier to ask yourself "Can I find discontinuities in f(x) or f'(x)?".
 
ultima9999 said:
I'm not sure where to start with this question. If a limit was given, I could solve it but without it given, I am completely lost...

State on which intervals the function [tex]f[/tex] defined by [tex]f(x) = \left\{\begin{array}{cc}|x + 1|,&x < 0\\x^2 + 1,&x \geq 0\end{array}\right.[/tex] is:

i) continuous
ii) differentiable
Find the derivative [tex]f'(x)[/tex] at all points where the function is differentiable.
Roughly speaking, a continuous function is a function whose graph is unbroken, i.e it has no "holes" or "jumps" (e.g the parabola y = x2 is everywhere continuous). Your function has absolute value in it, which makes it a little bit confusing when you look at it. So why don't we just break up the function to be:
[tex]f(x) = \left\{ \begin{array}{ll} - x - 1 & x < -1 \\ x + 1 & -1 \leq x < 0 \\ x ^ 2 + 1 & x \geq 0 \end{array} \right.[/tex]
Can you follow me?
Now the function is continuous on the interval [tex](- \infty , -1 )[/tex], and [tex](-1, 0 )[/tex], and [tex](0 , + \infty)[/tex] right? Do you know why?
Now to see if the function is continuous at -1, we simply check if:
[tex]\lim_{x \rightarrow -1 ^ -} f(x) = \lim_{x \rightarrow -1 ^ +} f(x) = f(-1)[/tex]. If the equations above hold, then the function is continuous at x = -1. Just do the same to see if the function is continuous at x = 0.
To see on which interval is he function f(x) differentiable, first you can try to find f'(x), then find where f(x) is indifferentiable (i.e where f'(x) is discontinuous or undefined). Then we can simply drop out the x values that make f(x) indifferentiable to obtain the interval on which f(x) is differentiable.
Can you go from here? :)
 
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