Question

the graph of a function $f(x)$ is shown below. which of the following could be the graph of $f(x)$?

Explanation:

Step1: Analyze the original function's graph

The graph of \( f(x) \) is symmetric about the \( y \)-axis (even function), has two local maxima (one on the left of \( y \)-axis, one on the right) and a minimum at \( x = 0 \).

Step2: Recall the relationship between \( f(x) \) and \( f'(x) \)

The derivative \( f'(x) \) represents the slope of the tangent to \( f(x) \) at any point \( x \).
- For \( f(x) \), when \( x < 0 \) (left of \( y \)-axis), the function first increases (slope positive), reaches a local maximum (slope zero), then decreases (slope negative) until \( x = 0 \) (where slope is zero at the minimum).
- When \( x > 0 \) (right of \( y \)-axis), due to symmetry, the function first increases (slope positive), reaches a local maximum (slope zero), then decreases (slope negative).

Step3: Analyze the options

- Option 1 (the first red graph, a downward - opening parabola symmetric about \( y \)-axis): This graph is symmetric about \( y \)-axis. For \( x < 0 \), it starts from below, rises to a maximum at \( x = 0 \), then falls. But our analysis of \( f'(x) \) for \( x < 0 \) should be positive - zero - negative, and for \( x>0 \) positive - zero - negative. Wait, no, let's re - evaluate. Wait, the original \( f(x) \): when \( x < 0 \), as \( x \) increases from \( -\infty \) to the local max, \( f(x) \) is increasing (so \( f'(x)>0 \)), at local max \( f'(x) = 0 \), then from local max to \( x = 0 \), \( f(x) \) is decreasing (so \( f'(x)<0 \)). At \( x = 0 \), \( f(x) \) has a minimum, so the slope changes from negative (just left of 0) to positive (just right of 0), so \( f'(x) \) crosses zero at \( x = 0 \) from negative to positive? Wait, no, the original graph: at \( x = 0 \), the function has a minimum (a "valley" at \( x = 0 \)). So the slope of \( f(x) \) just left of 0 is negative (function decreasing towards 0), and just right of 0 is positive (function increasing from 0). So \( f'(x) \) at \( x = 0 \) is zero, and crosses from negative to positive. But the first option is a downward - opening parabola, which is positive on the left and right of 0, zero at 0? No, wait, no. Wait, maybe I made a mistake. Wait, the original \( f(x) \) is symmetric about \( y \)-axis. Let's consider the derivative of an even function. If \( f(x) \) is even, \( f(-x)=f(x) \), then \( f'(-x)\times(- 1)=f'(x) \), so \( f'(-x)=-f'(x) \), which means \( f'(x) \) is an odd function. Wait, that's a key point! If \( f(x) \) is even, \( f'(x) \) is odd. So \( f'(x) \) should be symmetric about the origin. Let's check the options:
- The second option (the graph with a wave, symmetric about origin? Let's see: the graph on the left of 0 is a downward curve, on the right is an upward curve, symmetric about origin (odd function). Wait, no, let's re - check the options. Wait, the first option is symmetric about \( y \)-axis (even), the second option: when \( x < 0 \), it's a curve that goes down then up, and when \( x>0 \), it goes up then down, symmetric about origin (odd). The third option is piece - wise constant, not matching. The fourth option is not symmetric about origin. Wait, but our earlier analysis of the slope: for \( f(x) \) (even function), \( f'(x) \) is odd. So \( f'(x) \) should satisfy \( f'(-x)=-f'(x) \). So the graph of \( f'(x) \) should be symmetric about the origin. Let's check the second option: the graph on the left of 0 (negative \( x \)): as \( x \) increases from \( -\infty \) to 0, the slope of \( f(x) \): when \( x \) is very negative, \( f(x) \) is increasing (so \( f'(x)>0 \)? No, wait the original \( f(x) \) graph: when \( x \) is very negative, the function comes from \( -\infty \) up to a local max, so it's increasing (slope positive), then decreases (slope negative) to \( x = 0 \). At \( x = 0 \), slope is zero. Then for \( x>0 \), it increases (slope positive) then decreases (slope negative). Wait, but if \( f(x) \) is even, \( f'(x) \) is odd. So \( f'(-x)=-f'(x) \). Let's take a point \( x=a>0 \), \( f'(a) \) should be equal to \( - f'(-a) \). So the graph of \( f'(x) \) should be symmetric about the origin. The second option (the graph with the wave) is symmetric about the origin. Let's check the first option: it's symmetric about \( y \)-axis (even), which can't be since \( f'(x) \) of an even \( f(x) \) is odd. The third option is piece - wise constant, not matching the slope changes. The fourth option is not symmetric about origin. Wait, maybe I messed up the even - odd derivative. Let's derive it: if \( f(-x)=f(x) \), then differentiating both sides with respect to \( x \): \( -f'(-x)=f'(x) \), so \( f'(-x)=-f'(x) \), so \( f'(x) \) is odd. So \( f'(x) \) must be odd, so symmetric about origin. The second option's graph: when \( x < 0 \), the curve is below the \( x \)-axis (negative) then above (positive)? No, wait the second option's graph: left of 0, it's a curve that goes down (negative) then up (positive), crossing zero at 0, and right of 0, it goes up (positive) then down (negative), crossing zero at 0. Wait, no, let's look at the original \( f(x) \) again. The original \( f(x) \) has two local maxima (left and right) and a minimum at 0. So the derivative \( f'(x) \) should have:
- For \( x < 0 \): starts with positive (function increasing), reaches 0 at local max, then becomes negative (function decreasing) until \( x = 0 \) (where \( f'(x)=0 \)).
- For \( x>0 \): starts with positive (function increasing), reaches 0 at local max, then becomes negative (function decreasing). But due to the odd property, \( f'(-x)=-f'(x) \), so when \( x < 0 \), \( f'(x) \) should be the negative of \( f'(-x) \) (where \( -x>0 \)). Wait, maybe the second option is the correct one. Wait, the first option is a downward - opening parabola (even function), which is wrong because \( f'(x) \) should be odd. The second option: let's see the shape. The original \( f(x) \) has two peaks (local maxima) and a valley at 0. So the derivative \( f'(x) \) will have:
- Two zeros (at the local maxima of \( f(x) \)) on the left of 0, one zero at 0, and two zeros on the right of 0? No, each local maximum of \( f(x) \) gives a zero of \( f'(x) \), and the minimum at 0 gives a zero of \( f'(x) \). So \( f'(x) \) has three zeros: two at the local maxima of \( f(x) \) (symmetric about 0) and one at \( x = 0 \). The second option's graph has a zero at \( x = 0 \) and two other zeros (symmetric about 0), and the shape: for \( x < 0 \), it goes from negative (wait no, when \( x \) is very negative, \( f(x) \) is increasing, so \( f'(x)>0 \)). Wait, I think I made a mistake in the even - odd derivative. Let's take a simple even function, say \( f(x)=x^{2} \), \( f'(x) = 2x \), which is odd. Another even function: \( f(x)=-x^{4}+2x^{2} \), \( f'(x)=- 4x^{3}+4x=-4x(x^{2} - 1) \), which is odd (since \( f'(-x)=4x(x^{2}-1)=-f'(x) \)). The graph of \( f'(x)=-4x(x^{2}-1) \) is a cubic curve, symmetric about the origin. The original \( f(x) \) in the problem is similar to \( f(x)=-x^{4}+2x^{2} \) in terms of having two local maxima and a local minimum at 0. So \( f'(x) \) should be a cubic - like curve (odd function) with zeros at \( x = - a,x = 0,x = a \) (where \( x=\pm a \) are the local maxima of \( f(x) \)). The second option (the graph with the wave) is a curve that is symmetric about the origin, has a zero at \( x = 0 \), and two other zeros (symmetric about 0), and the shape: for \( x < 0 \), it first goes down (negative) then up (positive)? No, wait when \( x \) is very negative, for \( f(x)=-x^{4}+2x^{2} \), \( f'(x)=-4x^{3}+4x \), when \( x\to-\infty \), \( - 4x^{3}\to+\infty \), \( 4x\to-\infty \), but the dominant term is \( - 4x^{3} \), so \( f'(x)\to+\infty \) (positive). At \( x=-1 \), \( f'(-1)=4 + (-4)=0 \). Between \( -\infty\) and \( - 1 \), \( f'(x)>0 \) (function increasing), between \( - 1\) and \( 0 \), \( f'(x)=-4x(x^{2}-1) \), when \( x\in(-1,0) \), \( x^{2}-1<0 \), \( x<0 \), so \( - 4x(x^{2}-1)<0 \) (function decreasing). At \( x = 0 \), \( f'(0)=0 \). Between \( 0\) and \( 1 \), \( x>0 \), \( x^{2}-1<0 \), so \( - 4x(x^{2}-1)>0 \) (function increasing), and for \( x > 1 \), \( x^{2}-1>0 \), \( x>0 \), so \( - 4x(x^{2}-1)<0 \) (function decreasing). So the graph of \( f'(x) \) for \( f(x)=-x^{4}+2x^{2} \) is positive when \( x < - 1 \), negative when \( - 1<x<0 \), positive when \( 0<x < 1 \), negative when \( x>1 \). Which matches the second option's graph (the one with the wave: left of 0, it goes from positive (far left) down to negative, crosses zero at some negative \( x \), then at \( x = 0 \) crosses zero again, then goes positive then negative on the right). The first option is a downward - opening parabola (even function), which is wrong. The third option is piece - wise constant, not matching the smooth slope changes. The fourth option is not symmetric about origin. So the second option (the graph with the wave - like curve symmetric about origin) is the correct one. Wait, but in the options, the second option is the one with the red curve that has a "valley" on the left and a "peak" on the right, symmetric about the origin.

Answer:

The second option (the graph with the wave - like curve symmetric about the origin, the one in the middle of the four options)