AP Calculus AB/BC. 3.2 Differentiability, p. 109 Day 1. To be differentiable, a function must be continuous and smooth. Derivatives will fail to exist at:. Cusp – slopes of the secant lines approach −∞ from one side and ∞ from the other side. Corner – one-sided derivatives differ.
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3.2 Differentiability, p. 109
Derivatives will fail to exist at:
Cusp – slopes of the secant lines approach −∞ from one side and ∞ from the other side.
Corner – one-sided derivatives differ.
Vertical tangent – the slopes of the secant lines approach −∞ or ∞ from both sides.
Discontinuity – One of the one-sided derivatives is nonexistent.
Most of the functions we study in calculus will be differentiable.
Example 1: The function below fails to be differentiable at x = 0. Tell whether the problem is a corner, a cusp, a verticaltangent, or a discontinuity.
Since we want to differentiate at x = 0, we know at x = 0, y = 1 by the definition.
But, tan−10 = 0. So, therefore the problem is discontinuity because there are two values of y at x = 0.
For small values of h, the difference quotient often gives a good approximation of f′(x).
The graphing calculator uses nDERIV to calculate the numerical derivative of f.
A better approximation is the symmetric difference quotient which the graphing calculator uses.
Compute the numerical derivative of f(x) = 4x – x2 at x = 0.
Use h = 0.001.
(x + h)2
4(x + h)
You must be able to calculate derivatives with the calculator and without.
So, you need to do them by hand when called for.
Remember that half the test is no calculator.
Find at x = 2.
8: nDeriv( x ^ 3, x, 2, 0.001)
Note: nDeriv( x ^ 3, x, 2, 0.001) is the same as
nDeriv( x ^ 3, x, 2).
The calculator may return an incorrect value if you evaluate a derivative at a point where the function is not differentiable.
nDeriv( 1/x, x, 0)
nDeriv( abs(x), x, 0)
−10 ≤x≤ 10
−10 ≤ y ≤ 10
What does the graph look like?
This looks like:
Use your calculator to evaluate:
3.2 Differentiability, p. 109
Theorem 1: Differentiability Implies Continuity
If f has a derivative at x = a, then f is continuous at x = a.
Since a function must be continuous to have a derivative, if it has a derivative then it is continuous.
Theorem 2: Intermediate Value Theorem for Derivatives
If a and b are any two points in an interval on which f is differentiable, then f ′ takes on every value between f′(a) and f′(b).
Example 2: Find all values of x for which the function is differentiable.
For this particular rational function, factor the denominator.
The zeros are at x = 5 and x = −1 which is where the function is undefined.
Since f(x) is a rational function, it is differentiable for all values of x in its domain. Therefore, f(x) is not differentiable at x = 5 or x = −1 since 5 and −1 are not in the domain of f(x).