MATH 12002 - CALCULUS I 5.3: Curve Sketching and the Natural - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.3: Curve Sketching and the Natural Exponential Function

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Example

Example

Let f (x) = x2ex. Determine intervals where f is increasing, intervals where f is decreasing, the location of all local maxima and minima, intervals where f is concave up, intervals where f is concave down, the location of all inflection points, and sketch the graph of f .

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Example

We need to determine the signs of f ′ and f ′′ for f (x) = x2ex. We have f ′(x) = 2xex + x2ex = ex(x2 + 2x) = exx(x + 2) and f ′′(x) = ex(x2 + 2x) + ex(2x + 2) = ex(x2 + 4x + 2). Recall that ex is never 0 or negative. So f ′(x) = 0 when x = 0 or x = −2, and f ′′(x) = 0 when x2 + 4x + 2 = 0. By the quadratic formula, this occurs when x = −4 ± √ 42 − 4 · 1 · 2 2 = −4 ± √ 8 2 = −4 ± 2 √ 2 2 = −2 ± √ 2. Thus f ′′(x) = 0 when x = −2 − √ 2 ≈ −3.4 or x = −2 + √ 2 ≈ −0.6, and f ′′(x) = ex[x − (−2 − √ 2)][x − (−2 + √ 2)].

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Example

f (x) = x2ex, f ′(x) = exx(x + 2), f ′′(x) = ex[x − (−2 − √ 2)][x − (−2 + √ 2)]

−2 − √ 2 −2 −2 + √ 2

ex x x + 2 f ′(x) ex

x − (−2 − √ 2) x − (−2 + √ 2)

f ′′(x)

Inc-Dec Concave Shape

+ + + + + − − − − + − − + + + + + − − + + + + + + − + + + + − − − + + + − − + + ✛ ✛ I I D D I ✲ ✲ U D D U U ✌ ✎ ☞✍ ✌

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Example

f (x) = x2ex, f ′(x) = exx(x + 2), f ′′(x) = ex[x − (−2 − √ 2)][x − (−2 + √ 2)]

−2 − √ 2 −2 −2 + √ 2 Inc-Dec Concave Shape

I I D D I U D D U U ✌ ✎ ☞✍ ✌

INF MAX INF MIN

f is increasing on (−∞, −2) ∪ (0, ∞), f is decreasing on (−2, 0); f has a local maximum at x = −2, f has a local minimum at x = 0. f is concave up on (−∞, −2 − √ 2) ∪ (−2 + √ 2, ∞), f is concave down on (−2 − √ 2, −2 + √ 2); f has inflection points at x = −2 − √ 2 and x = −2 + √ 2.

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Example

In order to sketch the graph of f , we will need to plot the points whose x coordinates are in the sign chart. We need to evaluate f (x) = x2ex at these points: f (−2 − √ 2) = (−2 − √ 2)2e−2−

√ 2 ≈ 0.38

f (−2) = (−2)2e−2 = 4 e2 ≈ 0.54 f (−2 + √ 2) = (−2 + √ 2)2e−2+

√ 2 ≈ 0.19

f (0) = 02e0 = 0 Note that (0, 0) is the only x- or y-intercept, and f (x) = x2ex is positive for all other values of x. Also, lim

x→+∞ x2ex = +∞ and

lim

x→−∞ x2ex = 0

(though we do not yet have the tools to verify the second limit). Hence y = 0 is a horizontal asymptote at −∞.

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