Take a look at these graphs. The numbered graphs show the gradient functions of ones with letters, but some are missing…
Try to pair up a function with its gradient function. Which features of the graphs will you use to justify your reasoning?
If a graph can’t be paired up with anything, try to sketch a graph that would complete its pair.
Following on from the groups of cards in the Suggestion, I now need to consider how these cards can be paired up.
First group
I notice graph A has a stationary point at \(x=0\) and its gradient is positive before and negative afterwards. Graph E has stationary points to the right and left of the origin.
- Does graph 5 match A or E? How can I tell?
- How could I use transformations to sketch the missing graph?
- How could I use features to sketch the missing graph?
Second group
Graph D has an asymptote along the vertical axis whereas graph F has a turning point at \(x=0\).
- What do I look for in the gradient function for each of these features?
- What transformation can I use to sketch the missing graph easily?
Third group
I can use the asymptotes to help find a first pair, and I can also think about transformations, since some of these graphs are transformations of each other.
- What feature can I use to choose between graphs 3 and 4 to match with graph C?
- How can I use transformations again to sketch the missing graph?
- Can I be certain what the missing graph should look like?
Once you have sorted the graphs into pairs of functions \(f(x)\) and \(f'(x)\) and sketched any missing graphs, try to sketch the graphs of the second derivatives.
Also try to sketch graphs of functions whose derivative is \(f(x).\) What do you notice?
I could continue to base my reasoning on transformations. Alternatively, I could consider the behaviour of the graphs. How does this affect the second derivative, \(f''(x)\), or the antiderivatives (the family of functions whose derivatives are \(f(x)\))? I’ll take the second approach and consider a few examples.
I’ll say graph A is \(y=f(x)\), so graph 5 is \(y=f'(x)\) and I can consider the gradient of graph 5 to help me sketch \(y=f''(x).\) The gradient of graph 5 is negative as it passes through the origin, so \(f''(0)\) must be negative. Graph 5 has alternating maxima and minima with sections of either positive or negative gradient in between, so \(y=f''(x)\) crosses the \(x\)-axis whenever graph 5 has a stationary point and alternates between being positive and negative. I notice that the \(x\)-intercepts of \(y=f''(x)\) correspond to points of inflection in graph A.
This type of thinking also helps me to sketch the antiderivatives of the function shown in graph A. What will these functions look like when graph A crosses the \(x\)-axis? What about when graph 5 crosses the \(x\)-axis?
In graphs F and 2, I notice that graph 2 also alternates between positive and negative gradient, but this time there are no stationary points. How must my approach to sketching the second derivative of graph F differ from how I sketched the second derivative of graph A? How does this reflect the differences between graphs F and A or graphs 2 and 5?
If I apply the same type of thinking to graphs B and 1, how could I use the other graphs to confirm my ideas?
My sketches of some second derivatives look very similar to the original functions. How could I tell if these are actually the same?
