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Numerade Educator



Problem 49 Hard Difficulty

(a) What are the domain and range of $ f $?
(b) What is the x-intercept of the graph of $ f $?
(c) Sketch the graph of $ f $.

$ f(x) = \ln x + 2 $


(a) The domain of $f(x)=\ln x+2$ is $x>0$ and the range is $\mathbb{R}$.
(b) $y=0 \Rightarrow \ln x+2=0 \Rightarrow \ln x=-2 \Rightarrow x=e^{-2}$
(c) We shift the graph of $y=\ln x$ two units upward.

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Video Transcript

Okay, so we're going to find the domain, the range, the X intercept and draw the graph of this function. And before we start, make sure you realize that there are invisible parentheses around the X, so the plus two is not inside the logger of them. It's outside it. Okay, So if we were looking at just the plain old Y equals natural log of X function, the basic one its domain would be X is greater than zero. Because you can't take the natural log of anything that zero or or negative Onley positives and its range would be all real numbers. In fact, if we just want to picture the graph, it looks something like this. That's your regular natural log. A graph. Okay, Now what are we doing to it when we had to? We're shifting the whole thing up to shifting the whole thing up to Does not in any way change the domain. So the domain is still X is greater than zero. And if you prefer, you can write that as zero to infinity. Shifting the whole thing up to also does not impact the range at all, because if it's going from negative infinity to infinity and you shifted up to It's still doing that so we can say all remembers. And if you prefer, you can say negative infinity to infinity. Next we went to calculate the X intercept so the X intercept would be the point where the y coordinate zero So we can say that zero equals natural log of X plus two. We can subtract two from both sides and we get negative. Two equals natural log of X. So what this really means is negative two equals log based to log based E Excuse me of X and if we rearrange that into its exponential form, we have e to the negative. Second power equals X. So if you want, you could write that as one over e squared instead of either the second power. So the X intercept is the point one over E squared comma zero. Okay, now let's graph this. So we got a head start on the graph because we looked at the graph of the natural log function and we know we're just shifting it up, too. And remember that a natural log function has a vertical Assen toted X equals zero and shifting the graph up to is not going to change that at all. We also learned in the first part that the X intercept is going to be one over e squared, comma zero so X intercept at one over Eastport comma zero. That's approximately 0.13 comma zero So we can put that point in place and then draw the sketch.