§ 1.5 Differentiability and Continuity

Section Outline Differentiability and Nondifferentiability Graphs Nondifferentiable at a Point Continuous and Discontinuous Functions

Examples of Nondifferentiability

Graphs Nondifferentiable at a Point EXAMPLE The tax that you pay to the federal government is a percentage of your taxable income, which is what remains of your gross income after you subtract your allowed deductions. In the year 2001, there were five rates or brackets for a single taxpayer, as shown below. So if you are single and your taxable income was less than $27,050 in 2001, then your tax is your taxable income times 15% (.15). The maximum amount of tax that you will pay on your income in this first bracket is 15% of $27,050 or (.15) x 27,050 = 4057.50 dollars. If your taxable income is more than $27,050 but less than $65,550, then your tax is $4057.50 plus 27.5% of the amount in excess of $27,050. So, for example, if your taxable income is $50,000, then your tax is $4057.50 + .275(50,000 – 27,050) = 4057.5 + .275 x 22,950 = 10,368.75 dollars. Let x denote your taxable income in 2001 and T(x) your tax for that year.

Graphs Nondifferentiable at a Point CONTINUED Find a formula for T(x) if x is not over $136,750. Plot the graph of T(x) for 0 ≤ x ≤ 136,750. SOLUTION (a) The function would be:

Graphs Nondifferentiable at a Point CONTINUED (b) The graph of T(x) is shown below. Notice that the graph has a corner when x = 27,050 and is therefore nondifferentiable at that point.

Continuous Functions Definition Example Continuous: A function f is continuous at x = a provided that, roughly speaking, its graph has no breaks (or gaps) as it passes through the point (a, f (a)). That is, f (x) is continuous at x = a provided that we can draw the graph through (a, f (a)) without lifting our pencil from the paper. Notice that f (x) is not continuous at x = 3 but it is continuous for every other value of x.

Differentiability & Continuity

Limits & Continuity In order for (1) to hold, three conditions must be satisfied. f (x) must be defined at x = a. must exist. The limit must have the value f (a). A function will fail to be continuous at x = a when any one of these conditions fails to hold.

Differentiability & Continuity EXAMPLE Determine whether the following function is continuous and/or differentiable at x = 1. SOLUTION To save ourselves from doing unnecessary work, let’s check to see if the function is differentiable at x = 1. If it is, Theorem I says that it is automatically continuous at that point (in which case we would not even need to check for continuity). Both pieces of the graph, f (x) = x + 2 and f (x) = 3x meet at x = 1, as seen to the right.

Differentiability & Continuity CONTINUED Upon looking at the graph, it appears that the function has a corner at x = 1 and is therefore nondifferentiable at x = 1. However, just looking at a graph is generally not the best way to make an informed decision. If we analyze the function itself at x = 1, namely the slope of the function to either side of x = 1, we find that to the left of x = 1, the slope of the function is 1 (f (x) = x + 2). To the right of x = 1, the slope of the function is 3 (f (x) = 3x). Therefore, in terms of the slope, there is not a smooth transition from the left-most piece of the graph to the right-most piece of the graph. Therefore, the function is not differentiable at x = 1. So what does that say about the continuity of the function at x = 1? Nothing! Theorem I only says that if the function is differentiable at a point then it is continuous at that same point. It says nothing about nondifferentiability implying anything about continuity. Therefore, we must conduct an investigation for the purpose of determining whether the function is continuous at x = 1 or not. Using the Limit Definition of Continuity, we will determine whether the function is continuous at x = 1 or not. We will investigate the three conditions associated with the definition.

Differentiability & Continuity CONTINUED Condition 1: The function is defined at x = 1 since f (1) = (1) + 2 = 3. Condition 2: We will show that the limit of the function at x = 1 exists. x 3x x x + 2 1.1 3.3 0.9 2.9 1.01 3.03 0.99 2.99 1.001 3.003 0.999 2.999 1.0001 3.0003 0.9999 2.9999 As x approaches 1, it appears that the function approaches 3 from both sides of x = 1. Therefore we say,

Differentiability & Continuity CONTINUED Condition 3: As discussed before (see condition 2), the value of is 3 which we saw is equal to f (1) (see condition 1). Therefore, the function is continuous at x = 1 (though not differentiable).