Question Video: Finding the Inflection Point of the Curve of a Polynomial Function | Nagwa Question Video: Finding the Inflection Point of the Curve of a Polynomial Function | Nagwa

Question Video: Finding the Inflection Point of the Curve of a Polynomial Function Mathematics • Third Year of Secondary School

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Find the inflection point of 𝑓(π‘₯) = π‘₯Β³ βˆ’ 12π‘₯ βˆ’ 1.

03:16

Video Transcript

Find the inflection point of 𝑓 of π‘₯ equals π‘₯ cubed minus 12π‘₯ minus one.

First of all, what are inflection points? These are points where a curve goes from having an increasing slope, also called concave up, to a decreasing slope, also called concave down, or vice versa. So how do we find these points? Well, remember that the first derivative of a curve tells us the slope of the curve, and the second derivative tells us if the slope is increasing or decreasing. Where the slope is increasing, the first derivative is increasing, and so the second derivative is positive. Where the slope is decreasing, the first derivative is decreasing. So the second derivative is negative.

So we’re looking for points where the second derivative goes from being positive to negative or negative to positive. In other words, we’re looking for those points where the second derivative is equal to zero. So let’s firstly find what the second derivative is. We’ll do this by firstly finding the first derivative 𝑓 prime of π‘₯. And we do this by differentiating 𝑓 of π‘₯ with respect to π‘₯. We’ll do this term by term, starting with the first term, that is, π‘₯ cubed. We can do this using the power rule for differentiation, which tells us that the derivative of π‘₯ to the 𝑛th power is 𝑛π‘₯ to the power of 𝑛 minus one. Essentially, this means we multiply by the power and then subtract one from the power. So π‘₯ cubed differentiates to three π‘₯ squared.

So now we differentiate the next term, which is 12π‘₯. So now we differentiate the next term negative 12π‘₯. This just differentiates to negative 12. And our last term negative one is just a constant, and constants differentiate to zero. So 𝑓 prime of π‘₯ is equal to three π‘₯ squared minus 12. To find 𝑓 double prime of π‘₯, we differentiate 𝑓 prime of π‘₯ with respect to π‘₯.

Let’s start with the term three π‘₯ squared. Using the power rule again, we find that three π‘₯ squared differentiates to six π‘₯. And as negative 12 is just a constant, this differentiates to zero. So 𝑓 double prime of π‘₯ is six π‘₯. Remember, we said we’re looking for the points where the second derivative is equal to zero. So we’re going to set six π‘₯ equal to zero. Solving for π‘₯, we find that π‘₯ must be equal to zero. So we know we have an inflection point at π‘₯ equals zero. That’s the only solution to six π‘₯ equals zero.

But let’s find the corresponding 𝑦-value. We can do this by substituting π‘₯ is equal to zero into our equation for 𝑓 of π‘₯. This gives us zero cubed minus 12 multiplied by zero minus one, which gives us negative one. So we can conclude that the inflection point for this curve is zero, negative one. Let’s now draw a very quick sketch of the graph of π‘₯ cubed minus 12π‘₯ minus one so we can check our answer.

Let’s mark on what we found for our point of inflection zero, negative one. And in fact, to the left of this point, we can see that the graph is concave down. And to the right of this point, we can see that the graph is concave up. So zero, negative one is definitely the point of inflection for this curve.

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