Analyzing a Graph In Exercises 35-38, analyze and sketch the graph of the function. Label any relative extrema, points of inflection, and asymptotes. f ( x ) = 1 2 − e − x

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To determine To graph: The function f ( x ) = 1 2 − e − x and label the relative extrema, points of inflection and asymptotes. Explanation Given Information: The provided function is f ( x ) = 1 2 − e − x . Graph: Consider the function, f ( x ) = 1 2 − e − x Evaluate the first derivative of the function, by applying product rule and constant multiple rule. f ′ ( x ) = d d x ( 1 2 − e − x ) = d d x ( 2 − e − x ) − 1 = − 1 ( 2 − e − x ) − 2 × d d x ( 2 − e − x ) × d d x ( − x ) = − 1 ( 2 − e − x ) − 2 × ( e − x ) Further solve as, f ′ ( x ) = − e − x ( 2 − e − x ) 2 Set the derivative equal to zero to compute the value of x . − e − x ( 2 − e − x ) 2 = 0 e − x = 0 The value of e − x cannot be zero, therefore the function has no critical points and therefore it has no extreme points or inflection points. The horizontal asymptote is basically a horizontal line in the graph of the function when x approaches to ± ∞ , the horizontal asymptote of exponential function f ( x ) = a b x + c always occur at y = c . Evaluate the horizontal asymptote of the function f ( x ) = 1 2 − e − x when x approaches to − ∞ lim x → − ∞ ( 1 2 − e − x ) = lim x → − ∞ ( 1 ) lim x → − ∞ ( 2 − e − x ) = 1 lim x → − ∞ ( 2 ) − lim x → − ∞ ( e − x ) = 1 2 − ∞ = 1 − ∞ Further solve as, lim x → − ∞ ( 1 2 − e − x ) = 0 Evaluate the horizontal asymptote of the function f ( x ) = 1 2 − e − x when x approaches to ∞ lim x → ∞ ( 1 2 − e − x ) = lim x → ∞ ( 1 ) lim x → ∞ ( 2 − e − x ) = 1 lim x → ∞ ( 2 ) − lim x → ∞ ( e − x ) = 1 2 − 0 = 1 2 The vertical asymptote is basically the y- axis line in the graph of the function when the curves approaches closer to y- axis

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Answer 1

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Calculus: An Applied Approach (Min...

10th Edition

Ron Larson

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ISBN: 9781305860919

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BuyFind

Calculus: An Applied Approach (Min...

10th Edition

Ron Larson

Publisher: Cengage Learning

ISBN: 9781305860919

Solutions

Chapter

Section

Answer 3

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Answer 4

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Answer 5

To determine

To graph: The function f(x)=12−e−x and label the relative extrema, points of inflection and asymptotes.

Explanation of Solution

Given Information:

The provided function is f(x)=12−e−x.

Graph:

Consider the function, f(x)=12−e−x

Evaluate the first derivative of the function, by applying product rule and constant multiple rule.

f′(x)=ddx(12−e−x)=ddx(2−e−x)−1=−1(2−e−x)−2×ddx(2−e−x)×ddx(−x)=−1(2−e−x)−2×(e−x)

Further solve as,

f′(x)=−e−x(2−e−x)2

Set the derivative equal to zero to compute the value of x.

−e−x(2−e−x)2=0e−x=0

The value of e−x cannot be zero, therefore the function has no critical points and therefore it has no extreme points or inflection points.

The horizontal asymptote is basically a horizontal line in the graph of the function when x approaches to ±∞, the horizontal asymptote of exponential function f(x)=abx+c always occur at y=c.

Evaluate the horizontal asymptote of the function f(x)=12−e−x when x approaches to −∞

limx→−∞(12−e−x)=limx→−∞(1)limx→−∞(2−e−x)=1limx→−∞(2)−limx→−∞(e−x)=12−∞=1−∞

Further solve as,

limx→−∞(12−e−x)=0

Evaluate the horizontal asymptote of the function f(x)=12−e−x when x approaches to ∞

limx→∞(12−e−x)=limx→∞(1)limx→∞(2−e−x)=1limx→∞(2)−limx→∞(e−x)=12−0=12

The vertical asymptote is basically the y-axis line in the graph of the function when the curves approaches closer to y-axis

Analyzing a Graph In Exercises 35-38, analyze and sketch the graph of the function. Label any relative extrema, points of inflection, and asymptotes. f ( x ) = 1 2 − e − x

Calculus: An Applied Approach (Min...

Solutions

Calculus: An Applied Approach (Min...

Analyzing a Graph In Exercises 35-38, analyze and sketch the graph of the function. Label any relative extrema, points of inflection, and asymptotes.

f ( x ) = 1 2 − e − x

Explanation of Solution

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