Q. Identify the graph of y = ln x + 1

Q: What is the function written explicitly?

The function is y=\ln x+1, the natural logarithm of x shifted up by 1.

Q: What is the domain of y=\ln x+1?

Domain: x>0. The logarithm is defined only for positive x.

Q: What is the range of y=\ln x+1?

Range: all real numbers, \mathbb{R}. As x varies over (0,\infty), y attains every real value.

Q: Does the graph have any asymptotes?

Yes. vertical asymptote at x=0. There is no horizontal asymptote; y increases without bound as x\to\infty.

Q: Where are the x- and y-intercepts?

x-intercept: solve \ln x+1=0 gives x=e^{-1}=1/e, point (1/e,0). No y-intercept because x=0 is not in the domain.

Q: How is this graph related to y=\ln x?

It is y=\ln x shifted upward by 1 unit. Every point (x,\ln x) moves to (x,\ln x+1).

Q: Is the function increasing or decreasing; what are its derivatives?

y'=\frac{1}{x}, positive for x>0, so the function is increasing on its domain. y''=-\frac{1}{x^2}<0, so it is concave down.

Q: What is the inverse function of y=\ln x+1?

Solve y=\ln x+1 for x: x=e^{\,y-1}. Thus the inverse is f^{-1}(x)=e^{\,x-1}.

Answer

Identify the graph of y = ln x + 1

The graph is the natural logarithm shifted up 1 unit: \(y=\ln x+1\).

Quick properties and explanation:
– Domain: \(x>0\).
– Range: \((-\infty,\infty)\).
– Vertical asymptote: \(x=0\).
– x-intercept: solve \(\ln x+1=0\) gives \(x=e^{-1}\) (point \(\bigl(\tfrac{1}{e},0\bigr)\)).
– Point on the curve: \((1,1)\).
– Increasing for \(x>0\) since \(y’=\dfrac{1}{x}>0\); concave down since \(y”=-\dfrac{1}{x^2}<0\).
– Behavior: as \(x\) approaches \(0\) from the right, \(y\to -\infty\); as \(x\to\infty\), \(y\to\infty\).

Final description: the curve of \(y=\ln x\) shifted upward by 1 unit, with domain \((0,\infty)\), range \((-\infty,\infty)\), vertical asymptote \(x=0\), and x-intercept at \(\bigl(\tfrac{1}{e},0\bigr)\).

Detailed Explanation

Graph of y = ln x + 1 — Step-by-step explanation

Recognize the parent function and transformation.The given function is a vertical translation of the natural logarithm. The parent function is \(y = \ln x\). The equation \(y = \ln x + 1\) is obtained by shifting the graph of \(y = \ln x\) upward by 1 unit.
Domain and range.For the natural logarithm, the domain is positive real numbers. Therefore the domain of \(y = \ln x + 1\) is
\(x > 0\).

The range of the natural logarithm is all real numbers, and a vertical shift does not change that, so the range is

\(\displaystyle (-\infty, \infty)\).
Vertical asymptote.The parent graph \(y = \ln x\) has a vertical asymptote at \(x = 0\). A vertical shift does not move vertical asymptotes, so \(y = \ln x + 1\) also has a vertical asymptote at
\(x = 0\).
Intercepts and key points.There is no y-intercept because the graph is not defined at \(x = 0\).
Find the x-intercept by solving \(\ln x + 1 = 0\).

\(\ln x = -1\)

Exponentiate both sides with base \(e\):

\(x = e^{-1} = \dfrac{1}{e} \approx 0.3679\).

So the x-intercept is the point \(\left(\dfrac{1}{e},\,0\right)\).

Find a couple of additional convenient points:
- At \(x = 1\): \(y = \ln 1 + 1 = 0 + 1 = 1\). Point \((1,1)\).
- At \(x = e\): \(y = \ln e + 1 = 1 + 1 = 2\). Point \((e,2)\).
Slope and concavity (shape).Compute the first derivative to check monotonicity:
\(\dfrac{dy}{dx} = \dfrac{d}{dx}(\ln x + 1) = \dfrac{1}{x}\).

For \(x>0\), \(\dfrac{1}{x} > 0\), so the function is increasing on its entire domain.

Compute the second derivative to check concavity:

\(\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^2}\).

For \(x>0\), \(-\dfrac{1}{x^2} < 0\), so the graph is concave down for all \(x>0\).
End behavior.As \(x\) approaches 0 from the right, \(\ln x\) goes to negative infinity, so \(y = \ln x + 1\) goes to negative infinity. In words: as \(x\) gets very small and positive, the graph falls without bound.
As \(x\) becomes large, \(\ln x\) increases without bound (but slowly), so \(y = \ln x + 1\) increases without bound as well.
How to sketch the graph (step-by-step instructions).
1. Draw the vertical asymptote line at \(x = 0\).
2. Plot the key points found: \(\left(\dfrac{1}{e},0\right)\), \((1,1)\), and \((e,2)\).
3. Remember the graph is increasing and concave down for all \(x>0\).
4. Starting just to the right of \(x = 0\), draw the curve descending steeply toward negative infinity, passing through \(\left(\dfrac{1}{e},0\right)\), then rising through \((1,1)\) and \((e,2)\), and continuing to increase slowly as \(x\) increases.
Summary description.The graph of \(y = \ln x + 1\) is the standard natural logarithm curve shifted upward by 1 unit. It has domain \(x>0\), range all real numbers, a vertical asymptote at \(x=0\), x-intercept at \(\left(\dfrac{1}{e},0\right)\), is increasing and concave down on its domain, and passes through \((1,1)\) and \((e,2)\).

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FAQs

What is the function written explicitly?

The function is \(y=\ln x+1\), the natural logarithm of \(x\) shifted up by 1.

What is the domain of \(y=\ln x+1\)?

Domain: \(x>0\). The logarithm is defined only for positive \(x\).

What is the range of \(y=\ln x+1\)?

Range: all real numbers, \(\mathbb{R}\). As \(x\) varies over \((0,\infty)\), \(y\) attains every real value.

Does the graph have any asymptotes?

Yes. vertical asymptote at \(x=0\). There is no horizontal asymptote; \(y\) increases without bound as \(x\to\infty\).

Where are the x- and y-intercepts?

x-intercept: solve \(\ln x+1=0\) gives \(x=e^{-1}=1/e\), point \((1/e,0)\). No y-intercept because \(x=0\) is not in the domain.

How is this graph related to \(y=\ln x\)?

It is \(y=\ln x\) shifted upward by 1 unit. Every point \((x,\ln x)\) moves to \((x,\ln x+1)\).

Is the function increasing or decreasing; what are its derivatives?

\(y'=\frac{1}{x}\), positive for \(x>0\), so the function is increasing on its domain. \(y''=-\frac{1}{x^2}<0\), so it is concave down.

What is the inverse function of \(y=\ln x+1\)?

Solve \(y=\ln x+1\) for \(x\): \(x=e^{\,y-1}\). Thus the inverse is \(f^{-1}(x)=e^{\,x-1}\).

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