Inflection points (algebraic) | AP Calculus AB | Khan Academy
Let G of x = 1/4 x^4 - 4x^3 + 24x^2. For what values of x does the graph of G have an inflection point or have a point of inflection?
So, let's just remind ourselves what a point of inflection is. A point of inflection is where we change our concavity, or you could say where our second derivative G prime of x switches signs, switches which is signs.
So, let's study our second derivative. In order to study our second derivative, let's find it. So we know that G of x is equal to 1/4 x^4 - 4x^3 + 24x^2.
Given that, let's now find G prime of x. G prime of x is going to be equal to—I’m just going to apply the power rule multiple times. 4 * 1/4 is just 1—I'm not going to write the 1 down—it's going to be 1 * x^(4-1), or x^3. Then, -3 * 4 is -12, so that's -12x^(3-1), or -12x^2. Finally, 2 * 24 is 48, so that's 48x^(2-1), or 48x.
I could just write that as 48x. So there you have it, I have our first derivative.
Now we want to find our second derivative. So G prime prime of x is just the derivative of the first derivative with respect to x, applying the power rule again: 3x^2 - 24x + 48.
Let's think about where this switches signs. This is a continuous function; it's going to be defined for all x's. So, the only potential candidates for where it could switch signs are when this thing equals zero. So, let's see where it equals zero.
Let's set it equal to 0: 3x^2 - 24x + 48 = 0. Let's see, everything is divisible by three, so let's divide everything by three, we get x^2 - 8x + 16 = 0.
Can I factor this? Yeah, this factors to (x - 4)(x - 4), or you could just view this as (x - 4)^2 = 0, so x - 4 = 0, or x = 4.
Thus, G prime prime of 4 = 0. Now let's see what's happening on either side of that point to check if we're actually switching signs or not.
Let me draw a number line here, so this is at 2, 3, 4, 5, and I could keep going. We know that something interesting is happening right over here. G prime prime of 4 is equal to zero.
So, let's think about what the second derivative is when we are less than four.
Let me just try G prime prime of 0 since that'll be easy to evaluate. G prime prime of 0 is just going to be equal to 48. So when we are less than four, our second derivative G prime is greater than zero.
So we're actually going to be concave upwards over this interval to the left of four.
Now, let's think about the right of four. So let me evaluate G prime prime of 10. G prime prime of 10 is equal to 3 * 10^2, which is 300, minus 24 * 10, which is -240, plus 48.
So this is 60: 300 - 240 is 60 + 48. This is equal to 108, so it’s still positive.
So, on either side of four, G prime prime of x is greater than zero. So even though the second derivative at x = 4 is equal to zero, on either side we are concave upwards.
The second derivative is positive, and that was the only potential candidate.
There are no values of x for which G has a point of inflection. X = 4 would have been a value of x at which G had a point of inflection if the second derivative had switched signs here, going from positive to negative or negative to positive.
But it's just staying positive to positive. The second derivative is positive; it just touches zero right here and then goes positive again.
So going back to the question: for what x values does the graph of G have a point of inflection? No x values! I'll put an exclamation mark there just for drama!