University:
Whitman College
Course:
MATH-124 | Introduction to Calculus
Academic year:

2021
Views:

23

Pages:

2
Author:

thitinnlc0

Worksheet, Section 3.1 (Local Extrema) Name: Date: Directions: You are encouraged to work in groups, but you don’t need to do so. We’ll turn in what we could finish at the end of class. 1. Let f (x) = ln(x2 + 1). Find the critical points and the inflection points for f . SOLUTION: f 0 (x) = 2x x2 + 1 f 00 (x) = −2(x2 − 1) (x2 + 1)2 Sign chart on f 00 (x): f 00 (x) − x < −1 − x>1 + −1 < x < 1 Therefore, the critical point is x = 0 and the points of inflection are x = ±1. 2. Let f (x) = 4x4 + 55x3 − 21x2 + 3. Find the inflection points (you’ll want to use a quadratic formula calculator online or using your calculator). f 00 (x) = 48x2 + 330x − 42 ⇒ f 00 (x) = 0 ⇒ x = −7, 1/8 Using a sign chart for f 00 , we see that the concavity changes at both points (so both points are points of inflection0) 3. Let f (x) = −2x3 + 3x2 + 1. Find the local extrema using the first and second derivative tests. SOLUTION: We find that f 0 (x) = −6x2 + 6x = −6x(x − 1), and f 00 (x) = −12x + 6. For the first derivative test, we make a sign chart on f 0 (x): f 0 (x) − x<0 + 01 From the sign chart, we have a local max at x = 0 and a local min at x = 1. For the second derivative test, f 00 (0) = 6 > 0 ⇒ Local min atx = 0 f 00 (1) = −6 < 0 ⇒ Local max atx = 1 4. Let g 0 (x) = e−x (x + 1)(x + 3) be the first derivative of a function g. Construct a sign chart and determine the local extrema for g using the first derivative test. g 0 (x) + x < −3 Local max at x = −3 and local min at x = −1. Turn Page Over... − −3 < x < −1 + x > −1 5. Let f (x) = 2 + 2x2 − x4 . Construct sign charts for the first and second derivatives to determine on which intervals f is increasing, decreasing, concave up and concave down. SOLUTION: First derivative (factored form): f 0 (x) = −4x(x2 − 1) = −4x(x + 1)(x − 1) g 0 (x) + x < −1 − −1 < x < 0 + 01 Conclusion: g is increasing for x < −1, or 0 < x < 1. The function g is decreasing for −1 < x < 0 and x > 1. Second derivative (factored form): f 00 (x) = −4(3x2 − 1). √ We note that g 00 (x) = 0 where x = ±1/ 3. g 0 (x) − √ x < −1/ 3 √ + √ −1/ 3 < x < 1/ 3 −√ x > 1/ 3 √ √ √ √ Conclusion: g is concave down if x < −1/ 3 or x > 1/ 3, and g is concave up if −1/ 3 < x < 1/ 3. 6. Let p be a function whose second derivative is p00 (x) = (x + 1)(x − 2)e−x . (a) Construct a second derivative sign chart for p and determine all inflection points of p. Straightforward, but recall that e−x is always a positive number (for any x). e−x (x + 1)(x − 2) + x < −1 − −1 < x < 2 + x>2 Conclusion: The function has inflection points at x = −1 and x = 2. √ (b) Suppose you know that x = 5−1 is a critical point of p. Does p have a local extremum at x (if 2 so, what is it)? SOLUTION: Using a calculator, x ≈ 0.6, and from√the sign chart, we see that p00 (0.61) < 0, therefore, the function has a local maximum at x = ( 5 − 1)/2. (c) If the point (2, 12e−2 ) lies on the graph of p(x), and p0 (2) = −5e−2 , find the equation of the tangent line to p at x = 2. Does the tangent line lie above the curve, below the curve, or neither at this value? (Why?) SOLUTION: The equation of the tangent line is y − 12e−2 = −5e−2 (x − 2). We also know that p00 (2) = 0, and the concavity changes from negative to positive. The line will be above the curve to the left of x = 2, and above the curve to the right of x = 2 (so overall, the answer is “neither”).

Math 124 Worksheet, Section 3.1 (Local Extrema)

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