Video Transcript
Use derivatives to identify which of the following is the graph of 𝑓 of 𝑥 is equal to negative 𝑥 plus one cubed multiplied by 𝑥 minus two.
In order to identify which is the correct graph of the function, we would usually begin by working out the 𝑥- and 𝑦-intercepts. We find the 𝑦-intercept by setting 𝑥 equal to zero, and we find the 𝑥-intercept by setting 𝑓 of 𝑥 equal to zero and solving the equation. Whilst we could use this method to eliminate some of our options, we are asked to use derivatives. We begin by recalling that the derivative of 𝑓 of 𝑥, written 𝑓 prime of 𝑥, is equal to zero at any turning or critical point. Each of our graphs has at least two of these, and these can be classified as local minimum points, local maximum points, or points of inflection. We will consider the difference between these later.
In order to differentiate our function, we will need to use the product rule for differentiation. This states that if 𝑓 of 𝑥 is the product of two terms 𝑢 and 𝑣, then 𝑓 prime of 𝑥 is equal to 𝑢𝑣 prime plus 𝑣𝑢 prime. In this question, we will let 𝑢 equal negative 𝑥 plus one cubed and 𝑣 equal 𝑥 minus two. Differentiating negative 𝑥 plus one cubed gives us negative three multiplied by 𝑥 plus one squared. We multiply the exponent by the parentheses, reduce the exponent by one, and then multiply by the derivative of the expression inside the parentheses. And in this case, differentiating the expression inside the parentheses gives us one. This is known as the chain rule of differentiation.
Next, we differentiate the expression 𝑥 minus two with respect to 𝑥. This is equal to one. The derivative of 𝑥 is one, and differentiating the constant gives us zero. Substituting these expressions into the product rule gives us negative 𝑥 plus one cubed minus three multiplied by 𝑥 plus one squared multiplied by 𝑥 minus two. At this stage, we could set our expression for 𝑓 prime of 𝑥 equal to zero. Alternatively, we could try to simplify the right-hand side first. One way to do this would be to take out a common factor of negative 𝑥 plus one squared. This means that 𝑓 prime of 𝑥 is equal to negative 𝑥 plus one all squared multiplied by 𝑥 plus one plus three 𝑥 minus six. We can then collect like terms inside the square brackets, giving us negative 𝑥 plus one all squared multiplied by four 𝑥 minus five.
We now have a simplified expression for the derivative of 𝑓 of 𝑥. Our next step is to set this derivative equal to zero. Since the product of two terms equals zero, one or both of the terms themselves must also equal zero. This means that either negative 𝑥 plus one all squared equals zero or four 𝑥 minus five equals zero. Multiplying through by negative one and then square rooting both sides, the first equation gives us 𝑥 plus one is equal to zero. We can then subtract one from both sides of this equation, giving us 𝑥 is equal to negative one. We therefore have a critical or turning point at 𝑥 is equal to negative one. Graphs (A), (C), and (E) all have a turning point at this value of 𝑥. Graph (A) has a point of inflection, graph (C) a local maximum, and graph (E) a local minimum.
As graphs (B) and (D) do not have a turning point at 𝑥 equals negative one, we can rule out these options. We can solve our second equation by adding five to both sides, giving us four 𝑥 is equal to five. Dividing through by four, we have 𝑥 is equal to five over four or 1.25. This means that we have a second turning point at this value of 𝑥. Graphs (C) and (D) have three turning points. However, we have shown that our function has only two turning points at 𝑥 is equal to negative one and 𝑥 is equal to five over four. We can therefore rule out options (C) and (E).
If we consider graph (A), we see that it does indeed have a turning point at 𝑥 is equal to five over four. This turning point is a local maximum. We can therefore conclude that the graph of 𝑓 of 𝑥 which is equal to negative 𝑥 plus one all cubed multiplied by 𝑥 minus two is graph (A).
Whilst it is not required in this question, we could go one stage further and identify whether the turning points are local maximums, local minimums, or points of inflection by considering the second derivative. We know that if the second derivative is less than zero, we have a maximum point. If the second derivative is greater than zero, i.e., positive, we have a minimum point. Finally, if the second derivative is equal to zero and the third derivative is not equal to zero, we have a point of inflection. Depending on the complexity of the function, we will sometimes be asked to find the second and third derivatives.
