### Video Transcript

Find the inflection point on the curve π¦ equals six π₯ multiplied by π₯ add one squared.

Inflection points occur where a curve goes from having an increasing slope to a decreasing slope or a decreasing slope to an increasing slope. Or we can say that these occur where a curve goes from being concave up to concave down or concave down to concave up. Because the first derivative of a curve tells us the slope, we can use the second derivative to work out where a curve is increasing and where a curve is decreasing. Where a curve is increasing, the second derivative is positive. And where the curve is decreasing, the second derivative is negative.

So to find points where a curve goes from increasing to decreasing or decreasing to increasing, weβre looking for where the second derivative goes from being positive to negative or negative to positive. In other words, weβre looking for those points where the second derivative is equal to zero.

So in order to find where the second derivative is equal to zero, weβre going to need to start by finding the first derivative. We know that π¦ equals six π₯ multiplied by π₯ add one squared. Itβs gonna be a little bit easier to find the first derivative if we start by distributing the parentheses. We can rewrite π¦ as six π₯ multiplied by π₯ add one multiplied by π₯ add one. And distributing the parentheses gives us that π¦ equals six π₯ multiplied by π₯ squared add two π₯ add one. And then multiplying six π₯ by each term in the parentheses, we have that π¦ equals six π₯ cubed add 12π₯ squared add six π₯.

So weβre going to use this to now find dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯. One rule thatβs going to help us to do this is the power rule for differentiation, which tells us that if we want to differentiate π₯ to the πth power, this is π multiplied by π₯ to the π minus one power. Essentially, this is telling us that we multiply by the power and then subtract one from the power. So six π₯ cubed differentiates to 18π₯ squared, 12π₯ squared differentiates to 24π₯, and six π₯ just differentiates to six. So dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯, is equal to 18π₯ squared add 24π₯ add six.

But weβre looking for the second derivative. So we need to differentiate this once more. We can again differentiate this term by term. This gives us 36π₯ add 24. Note that six is a constant, and constants differentiate to zero. Remember, weβre looking for where the second derivative is equal to zero. So weβre going to need to set our second derivative equal to zero. That is, 36π₯ add 24 equals zero.

Subtracting 24 from both sides gives us that 36π₯ is equal to negative 24. And dividing both sides by 36 gives us that π₯ equals negative 24 over 36. We can actually cancel this down by dividing the numerator and the denominator by 12. And this gives us that π₯ is equal to negative two over three.

So what weβve found here is that we have an inflection point where π₯ is equal to negative two over three. Because thatβs where the second derivative is zero, because it goes from either being positive to negative or negative to positive. Letβs now find the corresponding π¦-value for this point. We can do this by substituting π₯ equals negative two over three into our equation for π¦.

Letβs begin to work this out by simplifying these parentheses first, negative two over three add one. That just gives us one over three. And one over three squared is just one over nine because we can square the numerator and square the denominator. Now we can multiply together negative two over three with one over nine. Remember, we do this just by multiplying the numerators and multiplying the denominators. This gives us π¦ equals six multiplied by negative two over 27. Multiplying these together gives us negative 12 over 27. And we can then divide both the numerator and the denominator by three, to simplify the fraction to four over nine. So weβve got that π¦ equals negative four over nine.

So weβve found our inflection point to be negative two over three, negative four over nine. And a quick sketch of the curve π¦ equals six π₯ multiplied by π₯ add one squared confirms that negative two over three, negative four over nine is a point of inflection because here we can see the curve goes from being concave up to concave down.