Why these function graphs correspond to this derivative function graph?

Question

The task is: For each graph, select the corresponding graph of the derivative function.

We have two graphs:

Their corresponding derivative graph (according to the answer) is:

I was able to get the right answer. However, why is that so? Yes, their slopes (y-axis) at the C graph are 0's at x = 0, 2 as in the graphs 4 and 6. But aren't that all similarities they have?

For example, if we take a look at the C graph, x = 1, the slope is supposed to be ~-1.5. But the graphs 4 and 6 in point x = 1 have slopes ~ -0.8 (maybe -1, it is hard to say from these plots, but definitely not -1.5)

Answer 1

There are a few hints you can use, some are general and some are specific to polynomials (or other functions whose derivative is known).

1) When a function has a min, max, or saddle, its derivative will have a root.

2) When a function has an inflection point, its second derivative will have a root, so its derivative will have a min, max, or saddle.

3) The derivative of a cubic polynomial is a quadratic polynomial.

4) The derivative of a positive polynomial is positive, and similarly with negatives.

Using these four tips, we can tell that the derivative must be a positive quadratic polynomial with roots at 0 and 2, and a min at 1.

To determine the value at the minimum, we can estimate the slope at the inflection point. To do this, draw a tangent line through the inflection point, intersecting the x- and y- axes. If you divide the y-intercept by the negative of the x-intercept, you will get the slope. Based on my sketch, it seems like the y-intercept is about 2 1/3 = 7/3, while the x-intercept is about 1 4/5 = 9/5. Thus, the slope should be about -35/27, which is approximately -1.3, a reasonably close estimate to the min of figure C.

Answer 2

Let $f(x)$ and $g(x)$ denote the functions whose graphs are shown in Figures 4 and 6, respectively. It appears that $f(1)=2$ and $g(1)={1\over2}$. If they both have the same derivative, we have $g(x)=f(x)-{3\over2}$.

Now $f(x)$ looks like a cubic polynomial with a double root at $x=2$ and a single root at $x=-1$. If this is the case, then $f(x)$ is a multiple of $(x-2)^2(x+1)$. In order to have $f(0)=2$, the multiple must be

$$f(x)={1\over2}(x-2)^2(x+1)$$

in which case we have

$$f'(x)={1\over2}(2(x-2)(x+1)+(x-2)^2)={1\over2}(x-2)(2x+2+x-2)={3\over2}x(x-2)$$

so that $f'(1)=-{3\over2}$, which agrees with what's shown in Figure C. Note, this doesn't prove that $f(x)$ is a cubic polynomial, it just says that a cubic interpretation is consistent with all the data. The take-home message is that estimating slopes by visual inspection can be a tricky business. But be happy: your estimate of $-0.8$ or maybe $-1$ is well within an order of magnitude of the "true" value, which in some scientific settings is a real accomplishment!

Answer 3

• It's wrong to assume that these two plots are cubic equations. Yes, they probably are. But we don't KNOW that.

• It appears that plot 4 is just plot 6 plus 2.

• • Thus both plots will probably have the same derivative.

• For both plots, it appears that the local max occurs at $x=0$.

• • So the derivative is probably positive-decreasing on $(-\infty, 0)$ and $0$ at $x=0$.

• For both plots, it appears that the local min occurs at $x=2$.

• • So the derivative is probably positive-increasing on $(2, \infty)$ and $0$ at $x=2$.

• For both plots, it appears that the inflection point is at $x=1$.

• • So the second derivative is probably $0$ at $x=1$.
• • So the derivative is probably a concave-up function with a minimum at $x=1$.

You can argue that a picture alone is not enough information to conclude with certainty that both plots have the same derivative. The fact that they did not include a grid in the background indicates that you should spend more time on "shapes" than on numbers. Don't be concerned about measuring the slope of a tangent line. It's just too hard to be accurate with the information they have given you.