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Let 𝑓(𝑥) = ln((𝑒^𝑥 − 2)^2 + 1 ) (𝑏 is some parameter) (a) Determine the domain ...

Jan 23, 2024

Let 𝑓(𝑥) = ln((𝑒^𝑥 − 2)^2 + 1 ) (𝑏 is some parameter) (a) Determine the domain of 𝑓(𝑥), and whether 𝑓(𝑥) bounded (from above\below). Justify your claims. (b)Find the asymptotes of 𝑓(𝑥). (c) Determine for which values of 𝑏, 𝑓(𝑥) has a critical point, then find and classify this point. (d)Sketch a graph of 𝑓(𝑥)

Generated Graph

Solution by Steps

step 1

To determine the domain of

f(x) = \ln((e^x - 2)^2 + 1)

, we need to find the values of

x

for which the argument of the logarithm is positive

step 2

The argument of the logarithm,

(e^x - 2)^2 + 1

, is always positive because a square of a real number is non-negative and adding 1 ensures it's strictly positive

step 3

Therefore, the domain of

f(x)

is all real numbers,

\mathbb{R}

step 4

To determine if

f(x)

is bounded, we observe that the logarithm function grows without bound as its argument grows larger

step 5

Since

(e^x - 2)^2 + 1

grows without bound as

x

increases,

f(x)

is not bounded above

step 6

However,

f(x)

is bounded below because the logarithm of a positive number is always greater than the logarithm of 1, which is 0

Answer

The domain of

f(x)

is all real numbers,

\mathbb{R}

, and

f(x)

is unbounded above and bounded below.

Key Concept

Domain of a Logarithmic Function

Explanation

The domain of a logarithmic function includes all real numbers for which the argument of the logarithm is positive.

step 1

To find the horizontal asymptotes of

f(x)

, we examine the behavior of

f(x)

x

approaches infinity and negative infinity

step 2

x

approaches negative infinity,

e^x

approaches 0, so

(e^x - 2)^2

approaches 4, and

f(x)

approaches

\ln(4 + 1) = \ln(5)

step 3

x

approaches positive infinity,

(e^x - 2)^2

grows without bound, and so does

f(x)

, indicating no horizontal asymptote in this direction

Answer

The horizontal asymptote of

f(x)

y = \ln(5)

x

approaches negative infinity. There is no horizontal asymptote as

x

approaches positive infinity.

Key Concept

Horizontal Asymptotes of Logarithmic Functions

Explanation

Horizontal asymptotes are found by examining the limits of the function as

x

approaches positive or negative infinity.

step 1

To determine for which values of

b

f(x)

has a critical point, we find the derivative and set it equal to 0

step 2

The derivative of

f(x)

is given by

\frac{d}{dx} \ln((e^x - 2)^2 + 1)

step 3

Setting the derivative equal to 0 and solving for

x

gives

x = \ln(2)

step 4

To classify the critical point, we would examine the second derivative or use the first derivative test

step 5

Since the derivative changes from positive to negative at

x = \ln(2)

, this is a local maximum

Answer

For

b = \ln(2)

f(x)

has a critical point, which is a local maximum.

Key Concept

Critical Points of a Function

Explanation

Critical points occur where the derivative is zero or undefined, and they are classified by examining the sign of the derivative around these points.

step 1

To sketch a graph of

f(x)

, we use the domain, asymptotes, and critical points found in previous steps

step 2

The domain is all real numbers, so the graph extends infinitely in the

x

-direction

step 3

The horizontal asymptote at

y = \ln(5)

x

approaches negative infinity is plotted

step 4

The critical point at

x = \ln(2)

is marked, indicating a local maximum

step 5

The graph is drawn to increase from the horizontal asymptote, reach a peak at the critical point, and then continue to rise without bound

Answer

The graph of

f(x)

extends over all real numbers, has a horizontal asymptote at

y = \ln(5)

, a local maximum at

x = \ln(2)

, and rises without bound as

x

increases.

Key Concept

Graphing Logarithmic Functions

Explanation

The graph of a logarithmic function is determined by its domain, asymptotes, and critical points.