# Continuous functions, discontinuous functions, rules of differentiation.

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1. Prove (using the defi nition of continuity), that f(x) = x^2 - 4 is a continuous function.
2. Give an example of a discontinuous function and explain why it is discontinuous.
3. Find the average rate of change of f(x) = x^3 from x1 = 1 to x2 = 3.
4. Find the derivative of f(x) = x^2 at the point x0 = 2 using the de nition of the derivative.
5. Using the rules of di erentiation, find the derivative of f(x) = 5x^4 + 3x.
6. Using the rules of di erentiation, fi nd the derivative of f(x) = x^3 e^x^2
7. Using either the First or Second Derivative Test, nd all maxima and minima of f(x) = x^3 + 2x^2 + 1.

Borys_S
School: UCLA

1. π(π₯) = π₯ 2 β 4 is a continuous function for any π₯.
Proof. Consider |π(π₯) β π(π₯1 )| = |π₯ 2 β π₯12 | = |π₯ β π₯1 ||π₯ + π₯1 | =
= |π₯ β π₯1 |(|(π₯1 β π₯) + 2π₯|) β€ |π₯ β π₯1 |(|π₯1 β π₯| + 2|π₯|).
We need to choose πΏ(π, π₯) > 0 for any given π₯ and π > 0 such that |π₯ β π₯1 | < πΏ(π, π₯)
guarantees that |π(π₯) β π(π₯1 )| < π.
Suppose |π₯| < π (such π exists for any π₯). We can always choose πΏ < 1.
π

2π+1

This way if we choose πΏ(π, π₯) = min (1, 2π+2) weβll get |π(π₯) β π(π₯1 )| β€ π 2π+2 < π, β

0, π₯ β€ 0
2. Consider a function π(π₯) = {
.
1, π₯ >...

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Anonymous
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