AP Calculus AB Chapter IV Lesson B Curve Sketching
local maxima Absolute maximum F I A B E G C J Absolute H K minimum D local minima
Summary of trip along curve critical points occur where the derivative is zero or undefined all local (relative) maxes and mins occur at the critical points; however, not all critical points are necessarily extrema function is said to be increasing where derivative is positive, and decreasing where derivative is negative Back to Graph...
function is concave up where it looks like a cup [or smile]; it is concave down where it looks like a frown inflection points are where the curve changes from concave up to concave down or visa versa; these points are the steepest or least steep points in immediate vicinity Back to Graph...
If f Increasing / Decreasing 0 x on an interval, then f is increasing on that interval If f 0 x on an interval, then f is decreasing on that interval If f 0 x on an interval, then f is constant on that interval
example 1 Determine all intervals where the function is increasing or decreasing: f x x 2 3 12 set f 3 f x x 12x 1 x 0, solve 2 3x 12 0 x 2 4 0 x x 2 2 0 critical numbers x = 2, -2
Sign Graph + + + + + + f 3 15 f 0 12 f 4 36 Solution increasing, 2 and 2, decreasing 2,2
Finding local extrema (max / min) Only place that a derivative can change sign is at the critical points Fermat s Theorem - All relative extrema of a function will be critical points The points where the derivative is zero or where the derivative does not exist are the points where local maxima and minima can occur (x-values are critical points)
Finding local extrema Not every critical point will be an extrema
First Derivative Test Suppose c is a critical number of a continuous function f If f changes from positive to negative at c, then f has a local maximum at c. If f changes from negative to positive at c, then f has a local minimum at c. If f does not change sign at c, then the function f has no local max or local min at c.
Before you conclude that a change of sign - on a sign graph - indicates a local extreme point, make sure that the original function is defined there. Warning!
example 2 Find the intervals on which the function is increasing or decreasing, and find any local extrema (explanation must be included for extrema). 4 2 f x 2x 4x 1 + + + + + + f 2 48 f.5 3 f.5 3 f 3 192
Solutions increasing: 1,0 and 1, decreasing:, 1 and 0,1 local max 0,1 local min 1, 1 and 1, 1
Note: On the Ap Exam, it is not sufficient to simply draw a sign graph and use it to show a local min or max. You must provide a written explanation! Example There is a local max at (0,1) because f (x) > 0 for all x in the interval (-1,0) and f (x) < 0 for all x in the interval (0,1). The explanation for the minimum would be similar.
Concavity (a) described as a small hill (concave down) or a valley (concave up) (b) the curve lies below the tangents where it is concave down and lies above the tangents where it is concave up
c. inflection point - a point where the function is continuous and the concavity of the point changes» find by setting the second derivative equal to zero, and then solving» possible inflection points exist where the second derivative is zero or the second derivative does not exist The following information relates the second derivative and concavity
If f (x) > 0 for all x in some interval then f(x) is concave up on that interval. If f (x) < 0 for all x in some interval then f(x) is concave down on that interval.
example3 For the function, find any inflection points and intervals where the function is concave up or concave down. 4 3 2 f x 3x 4x 12x 5 1 st Derivative 3 2 critical numbers 1, 0, 2 f x 12x 12x 24x
f x 36x 24x 24 2 nd Derivative 2 inflection points.549 and 1.215 + + + + + + f 2 168 f 0 24 f 2 72
inflection Solution points.549,2.317 and 1.215, 13.351 concavity concave up:,.549 and 1.215, concave down:.549,1.215
Second Derivative Test Suppose that c is a critical point of f(x) and f (c) = 0 (i) If f (c) < 0, then c is a local maximum (ii) If f (c) > 0, then c is a local minimum (iii) If f (c) = 0, then c can be anything - local min, local max, or neither
In proving a relative minimum or maximum, it is never enough to show that the derivative is zero at that point. Suppose the 2 nd derivative of a function is: 2 f x 6x 24 critical numbers are 2 and -2
Solutions To prove that there is a local maximum at x = -2 f > 0 for x = -3 and c = -2 ; curve is increasing f < 0 for x = 0 and c = -2 ; curve is decreasing Therefore, there is a local maximum at x = -2
example 4 For the function, (a) find the intervals of increase and decrease, (b) find any local minimums or maximums, and (c) find the intervals of concavity along with inflection points. Then, make a sketch of the graph. f x x 4 6x 2
+ + + + + + f 3 72 f 1 8 f 1 8 f 2 8
f 2 36 f 0 12 f 3 96
example 6 For the following function, find (i) any relative extrema (and define them as you would on the AP Exam), (ii) find any inflection points along with concavity, and (iii) use the Second Derivative Test - where possible - to classify the critical points 2 f x x 6 x 3 Solutions: ( i) 3.6,4.5 6,0 rel. max rel. min ii 7. 2,8.1 only inflection pt. concave... 6,7.2
End of Lesson B Complete practice problems in notes
AP Calculus AB Unit V, Lesson C SketchingCurves / Mean Value Thorem
Curve Sketching There will be three (3) types of problems (1) Given a function, find various characteristics, then sketch (2) Given some information about several points on the function or its derivatives, sketch the curve (3) Shown f or f, sketch the graph of the function, f
Ex7
Websites for additional examples on curve sketching, given the derivative of a function: Curve Sketching -- Concavity Curve Matching - derivatives
example 8 Sketch the graph of a function that satisfies all of the given conditions: f 0 f 2 f 4 0 f x 0 if x 0 or 2 x 4 f x 0 if 0 x 2 or x 4 f x 0 if 1 x 3 ; f x 0 if x 1 or x 3 Solution
Solution 36
Summary of Curve Sketching 1. Leading Coefficient Test 2. Domain 3. Intercepts [not always necessary] 4. Asymptotes [horizontal & vertical] 5. Calculate First Derivative [find critical numbers; also, any points where f is undefined
6. Calculate Second Derivative [find possible inflection points along with any points where f is undefined] 7. Intervals of Increase or Decrease [1 st derivative sign graph] 8. Local Extrema [maxima & minima] use First or Second Derivative Tests 9. Concavity and Points of Inflection [2 nd derivative sign graph] 10. Sketch the Curve If additional accuracy is needed near any point, compute the value of the derivative there; tangent indicates direction curve proceeds.
example1 Sketch a graph of the function: f x x 2x 2 2 1
Sketch a graph of the function: yx 4 4x 3
Mean Value Theorem Mean Value Theorem for Derivatives If a function f(x) is continuous and differentiable on the closed interval [a,b], then there exists a point c between a and b such that f c f b b f a a. A point c is guaranteed to exist such that the derivative, f (c), is equal to the slope of the secant line for the interval [a,b]. a
example 2 2 x Given f x 1 on the interval 1,2 ; M.V.T states that at some x-value between 1 and 2, the slope of the tangent line will be equal to the slope of the secant line.
example 3 Given the function f x 3x 5x 1 on the interval 2,5, find the value of c that satisfies the Mean Value Theorem. 2
Rolle s Theorem Let f be a function that satisfies the following three conditions: (i) f is continuous on the closed interval [a,b] (ii) f is differentiable on the open interval (a,b) (iii) f(a) = f(b) Then there exists a number c in (a,b) such that f (c) = 0
Explanation If the corresponding y-values for a and b are equal, then there is a point on the curve where the slope of the tangent is zero (since the slope of the secant would also be zero).
example 4 2 Given the function f x x 2x 3 on the interval Rolle's Theorem. 1,3 find the value for c that satisfies
example 5 f x x1 on 1,3 Write an equation for (a) the secant line PQ and (b) a tangent line to f in the interval (a,b) that is parallel to PQ, where P = [a, f(a)] and Q = [b,f(b)].