Question

the graph of $f$, the derivative of the function $f$, is shown in the figure above.

2 mark for review
which of the following statements must be true?
i. $f$ is continuous on the open interval $(a, b)$.
ii. $f$ is decreasing on the open interval $(a, b)$.
iii. the graph of $f$ is concave down on the open interval $(a, b)$.

a i only
b i and ii only
c i and iii only
d ii and iii only

Explanation:

Step1: Analyze Statement I

If \( f' \) exists on \( (a,b) \), then \( f \) is differentiable on \( (a,b) \). Differentiable functions are continuous, so \( f \) is continuous on \( (a,b) \). So Statement I is true.

Step2: Analyze Statement II

To determine if \( f \) is decreasing, we check \( f'(x) \). The graph of \( f'(x) \): for \( x \) in \( (a,b) \), when \( x>0 \) (assuming the intersection with origin, but looking at the graph, \( f'(x) \) is positive when \( x \) is to the right of the origin? Wait, no, the graph of \( f'(x) \): let's see, the curve \( y = f'(x) \) crosses the origin? Wait, the graph: above the x - axis (positive \( y \)) when \( x \) is greater than some point, and below (negative \( y \)) when \( x \) is less? Wait, no, the graph of \( f'(x) \): the right - hand part (near \( x = a \)) is above the x - axis (positive \( f'(x) \)) and the left - hand part (near \( x = b \)) is below? Wait, no, the x - axis has \( a \) on the positive side and \( b \) on the negative? Wait, the coordinate system: \( O \) is the origin, \( a \) is on the positive x - axis, \( b \) on the negative. The graph of \( f'(x) \): when \( x\in(a,b) \), let's see the sign of \( f'(x) \). The graph crosses the origin. For \( x > 0 \) (near \( a \)), \( f'(x)>0 \), for \( x < 0 \) (near \( b \)), \( f'(x)<0 \). But the interval is \( (a,b) \), which is from \( a \) (positive x) to \( b \) (negative x). Wait, no, maybe \( a \) is on the left and \( b \) on the right? Wait, the standard coordinate system: x - axis, \( a \) and \( b \) are points, with \( a \) to the left of \( O \) and \( b \) to the right? Wait, the graph: the curve \( y = f'(x) \) is increasing? No, the curve is concave? Wait, no, the graph of \( f'(x) \): let's re - examine. The key is: if \( f'(x) \) is positive, \( f \) is increasing; if \( f'(x) \) is negative, \( f \) is decreasing. Looking at the graph of \( f'(x) \): in the interval \( (a,b) \), is \( f'(x) \) always negative? No, because the graph of \( f'(x) \) is above the x - axis (positive) when \( x \) is close to \( a \) (right - hand side) and below (negative) when \( x \) is close to \( b \) (left - hand side). Wait, no, maybe I got the direction wrong. Let's assume that the interval \( (a,b) \) is from \( a \) (left) to \( b \) (right), but the graph of \( f'(x) \): the curve \( y = f'(x) \) has \( f'(x)<0 \) for all \( x\in(a,b) \)? No, that's not the case. Wait, the graph of \( f'(x) \): the curve is below the x - axis (negative \( f'(x) \)) for all \( x\in(a,b) \)? Wait, no, the graph shows that \( f'(x) \) is negative in the interval \( (a,b) \)? Wait, no, the right - most part (near \( a \)) is above the x - axis (positive \( f'(x) \)) and the left - most part (near \( b \)) is below (negative \( f'(x) \)). Wait, maybe the x - axis is labeled with \( a \) on the positive side and \( b \) on the negative side, so \( (a,b) \) is from positive x to negative x. So in the interval \( (a,b) \), when \( x\in(a,0) \), \( f'(x)>0 \) (since the graph is above the x - axis), and when \( x\in(0,b) \), \( f'(x)<0 \). So \( f'(x) \) is not always negative on \( (a,b) \), so \( f \) is not decreasing on \( (a,b) \). So Statement II is false.

Step3: Analyze Statement III

To determine concavity of \( f \), we look at \( f''(x) \), the derivative of \( f'(x) \). The concavity of \( f \) is determined by the sign of \( f''(x) \). If \( f''(x)<0 \), \( f \) is concave down; if \( f''(x)>0 \), \( f \) is concave up. The graph of \( f'(x) \): is \( f'(x) \) decreasing? The graph of \( f'(x) \) is a curve that is concave? Wait, the slope of \( f'(x) \) (which is \( f''(x) \)): if the graph of \( f'(x) \) is decreasing (since the curve is going from above the x - axis to below, and its slope is negative), then \( f''(x)<0 \) for all \( x\in(a,b) \). So \( f \) is concave down on \( (a,b) \). So Statement III is true.

Answer:

C. I and III only