Video: Finding the Intervals of Upward and Downward Concavity of a Polynomial Function

Find the intervals over which the graph of the function 𝑓(π‘₯) = π‘₯Β³ βˆ’ 3π‘₯Β² βˆ’ 7π‘₯ is convex downwards and convex upwards.

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Video Transcript

Find the intervals over which the graph of the function 𝑓 of π‘₯ is equal to π‘₯ cubed minus three π‘₯ squared minus seven π‘₯ is convex downwards and convex upwards.

The question gives us a function 𝑓 of π‘₯ which is a cubic polynomial. And we’re asked to find the intervals over which the graph of this function is convex downwards and convex upwards. Let’s start by recalling what it means for a function to be convex downward or convex upward on an interval.

For a twice differentiable function 𝑓, we say that 𝑓 is convex downward on some interval 𝐼 if its second derivative is positive on the entire interval 𝐼. And if the second derivative is negative on this interval 𝐼, then we say that 𝑓 is convex upward on this interval 𝐼.

In this question, we’re given a function 𝑓 of π‘₯ which is a polynomial, which means both the first and second derivative of 𝑓 of π‘₯ will be polynomials. And since polynomials are defined for all real values of π‘₯, 𝑓, 𝑓 prime, and 𝑓 double prime of π‘₯ will all be defined for all real values of π‘₯. So to find the intervals where our function is convex downwards and convex upwards, we just need to find an expression for 𝑓 double prime of π‘₯.

To find 𝑓 double prime of π‘₯, we need to differentiate 𝑓 of π‘₯ twice. Since 𝑓 of π‘₯ is a polynomial, we’ll differentiate this term by term by using the power rule for differentiation. We want to multiply by our exponents of π‘₯ and then reduce this exponent by one. This gives us 𝑓 prime of π‘₯ is equal to three π‘₯ squared minus six π‘₯ minus seven.

To find 𝑓 double prime of π‘₯, we need to differentiate 𝑓 prime of π‘₯ with respect to π‘₯. Again, this is a polynomial, so we’ll do this term by term by using the power rule for differentiation. This time, we get 𝑓 double prime of π‘₯ is equal to six π‘₯ minus six.

Remember, to find the intervals where our function is convex downward and convex upward, we need to check where our function 𝑓 double prime of π‘₯ changes from positive to negative. In this case, we can see that our function 𝑓 double prime of π‘₯ is a linear function. So we can just sketch this. So our function 𝑓 double prime of π‘₯ is a linear function with a slope of six and a 𝑦-intercept of negative six. So we can sketch the following graph of our line 𝑦 is equal to 𝑓 double prime of π‘₯.

Let’s now use this sketch to decide where our function 𝑓 of π‘₯ is convex downward. Our function will be convex downward whenever 𝑓 double prime of π‘₯ is greater than zero. From our sketch, we can see that our line 𝑦 is equal to 𝑓 double prime of π‘₯ is above the π‘₯-axis whenever π‘₯ is greater than the π‘₯-intercept. And we can say something similar if we look for the interval where our function 𝑓 of π‘₯ is convex upward. We want 𝑓 double prime of π‘₯ to be negative. And we can see that our function 𝑓 double prime of π‘₯ is below the π‘₯-axis whenever π‘₯ is less than our π‘₯-intercept.

So we need to find the value of this π‘₯-intercept. The π‘₯-intercept of our line will happen when six π‘₯ minus six is equal to zero. And we can solve this for π‘₯. We add six to both sides of our equation and then divide through by six. We get π‘₯ is equal to one. So we can add this to our sketch. Our π‘₯-intercept has a value of one.

So what have we shown? We’ve shown if π‘₯ is greater than one, then our second derivative of π‘₯ is positive. And if π‘₯ is less than one, then our second derivative of 𝑓 is negative. But remember, the second derivative being negative is the same as saying our function is convex up. And the second derivative being positive is the same as saying that our function is convex down. So in actual fact, we’ve shown if π‘₯ is less than one, then our function is convex up. And if π‘₯ is greater than one, then our function is convex down.

We could leave our answer like this. However, we’ll write this in set notation. Our function 𝑓 of π‘₯ is defined for all real values of π‘₯. So saying π‘₯ is less than one is the same as saying π‘₯ is in the open interval from negative ∞ to one. And we can say something similar for π‘₯ is greater than one. It’s the same as saying π‘₯ is in the open interval from one to ∞.

Therefore, we’ve shown that the graph of the function 𝑓 of π‘₯ is equal to π‘₯ cubed minus three π‘₯ squared minus seven π‘₯ is convex upward on the open interval from negative ∞ to one. And is convex downwards on the open interval from one to ∞.

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