Question

the graph of ( f ), the second derivative of the function ( f ), is shown above.

3 mark for review
which of the following could be the graph of ( f )?

a

b

c

d

Explanation:

Step1: Analyze \( f''(x) \) graph

The graph of \( f''(x) \) is a parabola opening upwards, with vertex at the origin and crossing the \( x \)-axis at \( x = 0 \) (since it's symmetric about the \( y \)-axis). So \( f''(x) \) is a quadratic function, likely \( f''(x)=ax^2 \) (with \( a>0 \)) because it's a parabola opening up with vertex at \( (0,0) \).

Step2: Integrate \( f''(x) \) to get \( f'(x) \)

Integrate \( f''(x) \). If \( f''(x)=ax^2 \), then \( f'(x)=\int ax^2dx=\frac{a}{3}x^3 + C \). Since the graph of \( f''(x) \) is symmetric and passes through the origin, the constant of integration \( C = 0 \) (assuming \( f'(0)=0 \) or from the symmetry). So \( f'(x) \) is a cubic function with odd degree, passing through the origin, and with a positive leading coefficient (since \( a>0 \)), so \( f'(x) \) will have a "S" - shaped curve, increasing for \( x>0 \) and decreasing for \( x<0 \) (or vice - versa depending on the coefficient, but since \( a>0 \), the derivative of \( f'(x) \) (which is \( f''(x) \)) is positive when \( |x|>0 \), so \( f'(x) \) is increasing when \( x>0 \) and decreasing when \( x<0 \), so the graph of \( f'(x) \) is a cubic curve that has a local maximum at \( x<0 \) and a local minimum at \( x>0 \), passing through the origin.

Step3: Integrate \( f'(x) \) to get \( f(x) \)

Integrate \( f'(x)=\frac{a}{3}x^3 \) to get \( f(x)=\int\frac{a}{3}x^3dx=\frac{a}{12}x^4 + D \). But we can also analyze the concavity and the shape from the derivatives. The second derivative \( f''(x)>0 \) for all \( x eq0 \) (since the parabola opens upwards), so \( f(x) \) is concave up for all \( x eq0 \), and has an inflection point at \( x = 0 \) (since \( f''(0)=0 \)). Now, looking at the options for \( f(x) \):
- Option A: The graph seems to have a shape that is concave down in some regions, which is not possible as \( f''(x)>0 \) everywhere except at \( x = 0 \).
- Option B: The graph of \( f(x) \) should be a function whose second derivative is always non - negative (except at \( x = 0 \)) and has an inflection point at \( x = 0 \). The cubic - like derivative analysis and the concavity (concave up everywhere except at the inflection point) match with a function that has a "U - like" shape but with an inflection at the origin? Wait, no. Wait, when \( f''(x) \) is a parabola opening up, \( f(x) \) is a quartic function (degree 4) with a minimum at \( x = 0 \)? Wait, maybe I made a mistake in the integration. Let's re - examine the \( f''(x) \) graph. The \( f''(x) \) graph is a parabola opening upwards, so \( f''(x)\geq0 \) for all \( x \), with \( f''(x) = 0 \) at \( x = 0 \). So \( f(x) \) is concave up for all \( x \), and has an inflection point at \( x = 0 \) (since \( f''(0)=0 \)). Now, the first derivative \( f'(x) \) is a cubic function (since the integral of a quadratic is a cubic). A cubic function with a positive leading coefficient (because \( f''(x) \) has a positive leading coefficient) will have a local maximum and a local minimum. Then \( f(x) \) (the integral of the cubic) will be a quartic function. Now, looking at the options, the graph of \( f(x) \) should be a function that is concave up everywhere (since \( f''(x)>0 \) for \( x eq0 \)) and has an inflection point at \( x = 0 \). The graph in option D: Wait, no. Wait, let's look at the options again. The \( f''(x) \) is a parabola opening up, so \( f(x) \) is a function whose second derivative is non - negative, so it's concave up everywhere (except at \( x = 0 \) where it's an inflection point). The graph of \( f(x) \) should be a function that has a minimum at \( x = 0 \)? Wait, no. If \( f''(x) \) is positive for \( x>0 \) and \( x<0 \) (since the parabola opens up), then \( f(x) \) is concave up on \( (-\infty,0) \) and \( (0,\infty) \), so it has an inflection point at \( x = 0 \). Now, the first derivative \( f'(x) \): when \( x<0 \), \( f''(x)>0 \), so \( f'(x) \) is increasing; when \( x>0 \), \( f''(x)>0 \), so \( f'(x) \) is increasing. Wait, that can't be. Wait, no, the graph of \( f''(x) \) is a parabola opening upwards with vertex at the origin. So for \( x<0 \), \( f''(x)>0 \) (since the parabola is above the \( x \)-axis for \( x eq0 \)), and for \( x>0 \), \( f''(x)>0 \). So \( f'(x) \) is a function that is always increasing (because \( f''(x)\geq0 \) for all \( x \)). Wait, I made a mistake earlier. If \( f''(x) \) is non - negative for all \( x \), then \( f'(x) \) is non - decreasing. The graph of \( f''(x) \) is a parabola opening upwards, so \( f''(x)=ax^2 \), \( a > 0 \). Then \( f'(x)=\frac{a}{3}x^3 + C \). If we assume \( f'(0)=0 \) (from the symmetry of \( f''(x) \) around the \( y \)-axis, maybe the function \( f(x) \) is even, so \( f'(x) \) is odd, so \( C = 0 \)). So \( f'(x)=\frac{a}{3}x^3 \), which is an odd function, increasing for all \( x \) (since the derivative of \( f'(x) \) is \( f''(x)=ax^2\geq0 \)). Then \( f(x)=\frac{a}{12}x^4+D \), which is an even function, with a minimum at \( x = 0 \) (since \( f'(x)<0 \) for \( x<0 \) and \( f'(x)>0 \) for \( x>0 \)), and concave up everywhere (since \( f''(x)>0 \) for \( x eq0 \) and \( f''(0)=0 \)). Now looking at the options, option D has a graph that is symmetric about the \( y \)-axis, has a minimum at \( x = 0 \), and is concave up everywhere (the shape of a quartic function \( y = x^4 \) - like, but with the correct concavity from the second derivative). The other options:
- Option A: The graph is not symmetric and has a shape that suggests concave down in some regions.
- Option B: The graph is not symmetric and has a cubic - like shape which would have \( f''(x) \) as a quadratic, but the concavity doesn't match.
- Option C: The graph is symmetric but has a shape that suggests concave down in some regions.
- Option D: The graph is symmetric about the \( y \)-axis, has a minimum at \( x = 0 \), and is concave up everywhere, which matches the fact that \( f''(x)>0 \) for all \( x eq0 \) and \( f''(0)=0 \) (inflection point at \( x = 0 \) for the second derivative, but for the function \( f(x) \), since \( f''(x) \) changes from 0 at \( x = 0 \) but is non - negative everywhere, \( f(x) \) has an inflection point? Wait, no. An inflection point is where the concavity changes. If \( f''(x)\geq0 \) for all \( x \) and \( f''(x)=0 \) only at \( x = 0 \), then the concavity does not change (it's concave up everywhere), so \( x = 0 \) is not an inflection point. Wait, I think I messed up the inflection point concept. An inflection point is where \( f''(x) \) changes sign. Since \( f''(x) \) is non - negative everywhere (the parabola opens upwards, so \( f''(x)\geq0 \) for all \( x \)), \( f''(x) \) never changes sign, so \( f(x) \) has no inflection point? But the graph of \( f''(x) \) is zero at \( x = 0 \). Wait, maybe the graph of \( f''(x) \) is a parabola that touches the \( x \)-axis at \( x = 0 \) (vertex at \( x = 0 \)), so \( f''(x)=ax^2 \), \( a>0 \). Then \( f'(x)=\frac{a}{3}x^3 + C \), and \( f(x)=\frac{a}{12}x^4+Dx + E \). If we assume \( f(x) \) is even (symmetric about \( y \)-axis), then \( D = 0 \). So \( f(x)=\frac{a}{12}x^4+E \), which is a U - shaped curve (like \( y = x^4 \)) with a minimum at \( x = 0 \), and concave up everywhere (since \( f''(x)=ax^2\geq0 \)). Among the options, option D has a graph that is symmetric about the \( y \)-axis, has a minimum at \( x = 0 \), and is concave up everywhere, which matches the behavior of \( f(x) \) given \( f''(x) \) is a parabola opening upwards.

Answer:

D (the graph in option D)