Question

How to do Piecewise functions

Answer 1

Here’s a practical how-to for working with piecewise functions (definition, evaluating, graphing, solving, continuity, derivatives, integrals) plus a worked example.

What a piecewise function is
- A function defined by different formulas on different parts of its domain, e.g.
f(x) = { rule1(x) for x in region1; rule2(x) for x in region2; ... }
- Each “piece” applies only on its specified interval (pay attention to ≤ or < at boundaries).

How to evaluate
1. Identify which piece's domain contains the x-value.
2. Plug x into that piece’s formula only.
Example: if f(x)= { x^2 for x<1; 2x+1 for x≥1 }, then f(0.5)=0.5^2=0.25, f(1)=2(1)+1=3.

How to graph
- Graph each formula only on its specified interval.
- Use a filled dot for included endpoints (≤ or ≥) and an open dot for excluded endpoints (< or >).
- Don’t connect pieces across breaks unless the values match at the boundary.

Solving equations/inequalities
- Solve separately on each piece’s interval, then keep only solutions that lie in that interval.
Example: solve f(x)=2 for f above:
- For x<1: x^2=2 → x=±√2. Only x=-√2 satisfies x<1.
- For x≥1: 2x+1=2 → x=0.5, but 0.5<1 so discard.
→ Solution: x = -√2.

Continuity at a boundary point x=a
- Compute left-hand limit L = lim_{x→a-} f(x) and right-hand limit R = lim_{x→a+} f(x).
- f is continuous at a iff L = R = f(a) (f(a) must be defined and equal to the common limit).
Tip: If L ≠ R, the function has a jump discontinuity.

Differentiability at a boundary point
- First the function must be continuous at the point. If not continuous, it’s not differentiable.
- If continuous, check left derivative and right derivative. The derivative exists there iff the two one-sided derivatives are equal.

Integration
- Break the integral at the piece boundaries and integrate each piece over the appropriate subinterval, then sum.

Worked example (putting it all together)
Let f(x)= { x^2 for x<1; 2x+1 for x≥1 }.
- Evaluate: f(0.5)=0.25, f(1)=3.
- Solve f(x)=2: as above, only x = -√2.
- Continuity at x=1: left limit = 1^2 = 1, right limit = 2(1)+1 = 3 → limits differ, so f is discontinuous at x=1.
- Differentiability at x=1: not differentiable because it’s not continuous (even though left derivative 2 and right derivative 2 happen to be equal).
- Integrate from -2 to 3: ∫_{-2}^{3} f(x) dx = ∫_{-2}^{1} x^2 dx + ∫_{1}^{3} (2x+1) dx = [x^3/3]_{-2}^{1} + [x^2+x]_{1}^{3} = 3 + 10 = 13.

Common pitfalls
- Forgetting to restrict solutions to the piece’s domain.
- Mixing up open/closed endpoints.
- Assuming continuity or differentiability at a boundary without checking both one-sided behaviors.

If you want, give me a specific piecewise function and I’ll evaluate, graph, test continuity/differentiability, or integrate it step-by-step.