MATH 12002 - CALCULUS I 3.3: Information from the Graph of the - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.3: Information from the Graph of the Derivative

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Summary of Information from the Derivative

We have previously derived the following relationships between a function f and its first and second derivatives f ′ and f ′′: f increasing ← → f ′ positive f decreasing ← → f ′ negative f local max/min ← → f ′ = 0 or DNE & changes sign f concave up ↔ f ′ increasing ↔ f ′′ positive f concave down ↔ f ′ decreasing ↔ f ′′ negative f inflection point ↔ f ′ max/min ↔ f ′′ = 0 or DNE & changes sign In particular, this shows that if we have the graph of f ′, we can easily determine where f is increasing or decreasing, concave up or concave down, and find x-coordinates of local maxima and minima and inflection points.

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Example 1

The graph of the DERIVATIVE f ′ of a function f is given below. Determine intervals where f is increasing and intervals where f is decreasing, the x-coordinates of all local maxima and minima, intervals where f is concave up and intervals where f is concave down, the x-coordinates of all inflection points.

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Example 1

To determine increasing/decreasing and local maxima/minima, we use the sign of f ′: f ′ is positive where its graph is above the x-axis, on (−5, 1) ∪ (5, 6); f ′ is negative where its graph is below the x-axis, on (1, 5); f ′ is zero where its graph crosses the x-axis, at x = 1 and x = 5. Therefore, f is increasing on (−5, 1) ∪ (5, 6) and f is decreasing on (1, 5). The critical numbers of f are x = 1 and x = 5. At x = 1, f changes from increasing to decreasing, so f has a local maximum at x = 1. At x = 5, f changes from decreasing to increasing, so f has a local minimum at x = 5.

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Example 1

To determine concavity and inflection points, we use the sign of the second derivative, which is determined by whether f ′ is increasing or decreasing: f ′ is increasing on (−5, −3) ∪ (4, 6); f ′ is decreasing on (−3, 4). Therefore, f is concave up on (−5, −3) ∪ (4, 6) and f is concave down on (−3, 4). At x = −3 and x = 4, f ′ has local extrema and f ′′ = 0. At both points, f changes concavity, so f has inflection points at x = −3 and x = 4.

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Example 2

The graph of the DERIVATIVE f ′ of a function f is given below. Determine intervals where f is increasing and intervals where f is decreasing, the x-coordinates of all local maxima and minima, intervals where f is concave up and intervals where f is concave down, the x-coordinates of all inflection points.

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Example 2

f ′ is positive on (−4, 1) ∪ (1, 5); f ′ is negative on (−5, −4) ∪ (5, 6); f ′ is zero at x = −4, x = 1, and x = 5 — the critical numbers of f . Therefore, f is increasing on (−4, 1) ∪ (1, 5) and decreasing on (−5, −4) ∪ (5, 6). At x = −4, f changes from decreasing to increasing, so f has a local minimum at x = −4. At x = 5, f changes from increasing to decreasing, so f has a local maximum at x = 5. At x = 1, f ′ does not change sign and f has no local extremum.

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Example 2

f ′ is increasing on (−5, −2) ∪ (1, 3); f ′ is decreasing on (−2, 1) ∪ (3, 6). Therefore, f is concave up on (−5, −2) ∪ (1, 3) and f is concave down on (−2, 1) ∪ (3, 6). At x = −2, x = 1, and x = 3, f ′ has local extrema and f ′′ = 0. At all three points, f changes concavity, so f has inflection points at x = −2, x = 1, and x = 3.

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