Question

question 5 (1 point)
an equation for the graph shown is most likely.

y
0
x

a) $y = x + 1 - \log x$

b) $y = -\sqrt{x} + 1$

c) $y = \log x + 1 - x$

d) $y = \log(\sqrt{x})$

Explanation:

Step1: Analyze Domain

The graph is defined for \( x \geq 0 \) (starts at \( x = 0 \)). Check domains:
- Option a: \( \log x \) requires \( x > 0 \), so \( x = 0 \) is undefined. Eliminate a.
- Option b: \( \sqrt{x} \) is defined for \( x \geq 0 \), but \( y = -\sqrt{x} + 1 \) is decreasing (since coefficient of \( \sqrt{x} \) is negative), but the graph is increasing. Eliminate b.
- Option c: \( \log x \) requires \( x > 0 \), \( x = 0 \) undefined. Eliminate c.
- Option d: \( \sqrt{x} \) is defined for \( x \geq 0 \), and \( \log(\sqrt{x})=\frac{1}{2}\log x \) (for \( x > 0 \)) and at \( x = 0 \), the limit as \( x \to 0^+ \) is \( -\infty \)? Wait, no—wait, the graph shown has a minimum at \( x = 0 \)? Wait, no, recheck. Wait, maybe I made a mistake. Wait, the graph in the image starts at \( (0,0) \) or near \( x=0 \), increasing. Wait, let's re - evaluate:

Wait, option a: \( y=x + 1-\log x \). The domain is \( x>0 \), but the graph in the problem seems to start at \( x = 0 \). Wait, maybe the graph is for \( x\geq0 \), and at \( x = 0 \), let's check the behavior as \( x\to0^+ \):

For option a: As \( x\to0^+ \), \( \log x\to-\infty \), so \( y=x + 1-\log x\to+\infty \), which doesn't match the graph starting near \( y = 0 \).

Option b: \( y=-\sqrt{x}+1 \). At \( x = 0 \), \( y = 1 \), and as \( x \) increases, \( y \) decreases (since \( -\sqrt{x} \) decreases), but the graph is increasing. So b is decreasing, eliminate.

Option c: \( y=\log x+1 - x \). As \( x\to0^+ \), \( \log x\to-\infty \), \( -x\to0 \), so \( y\to-\infty \), and as \( x\) increases, the derivative \( y'=\frac{1}{x}-1 \). For \( x>1 \), \( y'<0 \) (decreasing), so the function has a maximum, not increasing. Eliminate c.

Option d: \( y=\log(\sqrt{x})=\frac{1}{2}\log x \). Wait, at \( x = 1 \), \( y = 0 \); at \( x = 100 \), \( y=\frac{1}{2}\log100 = 1 \). Wait, no, that's too slow. Wait, maybe I misread the graph. Wait, the original graph—maybe the first analysis was wrong. Wait, let's check the sign of the functions:

Option b: \( y = -\sqrt{x}+1 \). The range: \( \sqrt{x}\geq0 \), so \( -\sqrt{x}\leq0 \), \( y\leq1 \). The graph in the problem is increasing and goes to \( +\infty \), so b is bounded above, eliminate.

Option a: \( y=x + 1-\log x \). The derivative \( y'=1-\frac{1}{x} \). For \( x>1 \), \( y'>0 \) (increasing); for \( 0<x<1 \), \( y'<0 \) (decreasing). So it has a minimum at \( x = 1 \), \( y=1 + 1-\log1=2 \). Wait, but the graph in the problem—maybe the graph is of \( y=x + 1-\log x \). Wait, at \( x = 1 \), \( y = 2 \); at \( x = e \), \( y=e + 1 - 1=e\approx2.718 \), increasing. At \( x\to0^+ \), \( y\to+\infty \), but maybe the graph is drawn with a minimum near \( x = 1 \)? Wait, maybe my initial domain analysis was wrong. Let's re - check the domain of option a: \( \log x \) is defined for \( x>0 \), so the graph can't include \( x = 0 \), but the given graph seems to start at \( x = 0 \). Wait, maybe the graph is for \( x>0 \) and the \( x = 0 \) is a limit. Let's check the behavior of each function:

Option a: \( y=x + 1-\log x \). The function is decreasing on \( (0,1) \) (since \( y'=1-\frac{1}{x}<0 \) when \( x<1 \)) and increasing on \( (1,+\infty) \) (since \( y'=1 - \frac{1}{x}>0 \) when \( x>1 \)), which matches a graph that has a minimum (a "U" - shaped part) and then increases, which the given graph (starting at a low point and increasing) could match.

Option b: Decreasing, as we saw, so no.

Option c: \( y=\log x+1 - x \). Derivative \( y'=\frac{1}{x}-1 \). Increasing on \( (0,1) \), decreasing on \( (1,+\infty) \), so it has a maximum, not matching the increasing after a point.

Option d: \( y=\log(\sqrt{x})=\frac{1}{2}\log x \). This is an increasing function, but at \( x = 1 \), \( y = 0 \); at \( x = 10 \), \( y=\frac{1}{2}\log10\approx0.5 \), which is a very slow increase, and as \( x\to0^+ \), \( y\to-\infty \), which doesn't match the graph starting near \( y = 0 \).

So the only function that has a minimum (decreasing then increasing) and is defined for \( x>0 \) (and the graph could be drawn with \( x>0 \)) is option a. Wait, but earlier I thought the domain was \( x\geq0 \), but maybe the graph is for \( x>0 \) (the origin is a limit). So the correct answer is a.

Step2: Confirm Behavior

- Option a: \( y=x + 1-\log x \). At \( x = 1 \), \( y=1 + 1-0 = 2 \). For \( x>1 \), \( y'=1-\frac{1}{x}>0 \) (increasing), for \( 0<x<1 \), \( y'=1 - \frac{1}{x}<0 \) (decreasing), so it has a minimum at \( x = 1 \), which matches the graph's shape (a curve that dips then rises).
- Option b: Decreasing, so no.
- Option c: Has a maximum, so no.
- Option d: Slow increasing, no minimum, so no.

Answer:

a) \( y = x + 1-\log x \)