Video: MATH-DIFF-INT-2018-S1-Q06

If 𝑓′(π‘₯) = βˆ’2π‘₯ + 6, which of the following statements is not true? a) The curve of the function 𝑓 is convex upwards over the interval ]βˆ’βˆž,∞[. b) The function 𝑓 has a local minimum value at π‘₯ = 3. c) The curve of the function 𝑓 has no inflection points. d) 𝑓(π‘₯) is decreasing over the interval ]3,∞[.

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Video Transcript

If 𝑓 dash of π‘₯ is equal to negative two π‘₯ plus six, which of the following statements is not true? Part a) the curve of the function 𝑓 is convex upwards over the interval of negative ∞ to ∞.

Here we note that the outward-facing brackets mean that this is an open interval, since negative and positive ∞ can never be included. For part a of this question, let us first recall the definition of convex upwards. A curve is said to be convex upwards when its gradient is decreasing. We should be familiar with the fact that when plotted in the π‘₯𝑦-plane, the value of 𝑓 of π‘₯ will give us the 𝑦-value of the curve. From this, it follows that the value of the first derivative, 𝑓 dash of π‘₯, will give us the gradient of the curve.

And, finally, it follows that the second derivative, 𝑓 double dash of π‘₯, will give us the rate of change of the gradient for any given point on the curve. For part a of the question, we are interested in the interval where the curve is convex upwards, which means the gradient is decreasing. This must mean that 𝑓 double dash of π‘₯ must be negative for this interval, or strictly less than zero.

To answer this part of the question, let us, therefore, find the second derivative of 𝑓 with respect to π‘₯, 𝑓 double dash of π‘₯. Now, the question has given us 𝑓 dash of π‘₯, the first derivative. And we can, therefore, find the second derivative by differentiating this. To move forward, we substitute in negative two π‘₯ plus six for 𝑓 dash of π‘₯. We then differentiate negative two π‘₯ plus six with respect to π‘₯, which gives us negative two. Here we have found that the value of 𝑓 double dash of π‘₯ is a constant and is equal to negative two. The value of 𝑓 double dash of π‘₯ has no π‘₯ term in the equation and, therefore, has no dependents on π‘₯ at all. This means that 𝑓 double dash of π‘₯ will be negative two regardless of the value that π‘₯ takes.

We can also know that 𝑓 double dash of π‘₯ is strictly less than zero, which means our curve is convex upwards. Since we’ve proved that 𝑓 double dash of π‘₯ is negative two everywhere, this means that our curve is convex upwards everywhere. This does indeed satisfy the condition that the curve must be convex upward over the entire open interval of negative ∞ to ∞. We have, therefore, proved that part a of the question is indeed true.

Part b) the function 𝑓 has a local minimum value at π‘₯ is equal to three. For this part of the question, the first you may recall is that a turning point, or a point of inflection, occurs when 𝑓 dash of π‘₯ is equal to zero. It would therefore follow to substitute in the value of π‘₯ equals three into the equation for 𝑓 dash of π‘₯ given by the question. The value of this is then given by negative two times three plus six, which, of course, simplifies to negative six plus six and is therefore equal to zero.

From this, we could, therefore, conclude that the curve does indeed have a turning point or a minimum or maximum value at π‘₯ is equal to three. We would then recall that when 𝑓 dash of π‘₯ is equal to zero, if 𝑓 double dash of π‘₯, the second derivative, is greater than zero, then we have a local minimum value. If 𝑓 double dash of π‘₯ is equal to zero, we have a point of inflection. And if 𝑓 double dash of π‘₯ is less than zero, we have a local maximum value.

Now, for part a of this question, we found that 𝑓 double dash of π‘₯ was a constant. And this was equal to negative two. Since this is the case, 𝑓 double dash of π‘₯ is always less than zero. And, therefore, we can only have local maximum values on this curve. In fact, we could’ve used this information to entirely skip the first few steps of this question. Since we know that our curve can never have any local minimum values, we know that part b of this question must not be true. We have now answered part b. And we can move on to part c of the question.

Part c) the curve of the function 𝑓 has no inflection points. Now, part c of this question can be answered using exactly the same logic as we used for part b. We know that the conditions for an inflection point are that 𝑓 double dash of π‘₯ must be equal to zero. We previously proved that 𝑓 double dash of π‘₯ is equal to negative two for all points. And, therefore, the curve of the function 𝑓 does indeed have no inflection points. We have, therefore, proved that part c of this question is indeed true. And the curve of the function 𝑓 has no inflection points.

Part d) 𝑓 of π‘₯ is decreasing over the interval between three and ∞. Again, we note that the outward-facing brackets denote that this is an open interval which does not include three or ∞. The first thing we can do for part d is to recognise that the open interval of three to ∞ is represented by π‘₯ being strictly greater than three. Any other values of π‘₯ would lie outside this interval. Let’s now find the values of π‘₯ over which 𝑓 of π‘₯ is decreasing. If 𝑓 of π‘₯ is decreasing, this must mean that the gradient of the curve of our function is negative. We, therefore, have that if 𝑓 of π‘₯ is decreasing, 𝑓 dash of π‘₯, representing the gradient, is strictly less than zero.

Now, the question has given us an equation for 𝑓 dash of π‘₯. And we can, therefore, substitute in minus two π‘₯ plus six into this inequality. We can now add two π‘₯ to both sides of the inequality and then divide by two. Doing so, we find that 𝑓 of π‘₯ is decreasing when π‘₯ is greater than three. If we flip around our inequality, we can more easily see that this is exactly the same as the interval stated by part d of the question. We, therefore, conclude that part d of the question is indeed true. And 𝑓 of π‘₯ is decreasing over the open interval of three to ∞.

We have now completed the question. And we have found that part a, c, and d are true. However, the statement in part b of the question is not true. The function does not have a local minimum value at π‘₯ is equal to three.

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