 So one of the more important things we can do with calculus is to graph functions. Now, this might seem to be a little bit of an archaic throwback after all. Do we not have graphing calculators and computer algebra systems? And the answer is, well, yes, actually we do. But one problem is that these are computers and they will only show us what we ask them to show us. For example, consider the graph of y equals x to the fourth e to the minus x. Now I can enter this into a graphing calculator and it will produce a graph, but not necessarily a graph that's useful. So it might produce a graph that looks like this. And the problem is I don't know whether this graph includes all of the important features of the function. In particular, there may be features of this graph that aren't shown here. So maybe I'll zoom out and take a bigger picture. Except now all of my important details have been crushed down to nothing at the center of the graph. So this graph isn't useful either. So let's try and use calculus to figure out what interval we should look at. Now the things we've looked at are critical points and inflection points. So let's go ahead and find those and then we'll construct an interval that includes all of these points. So first, let's find the critical points. We'll differentiate our function. Looking ahead, it'll turn out to be easiest if we factor these at least partially. And both terms have an e to the minus x, so let's remove that common factor. The critical points are going to be where the derivative does not exist or where the derivative is equal to zero. Since e to the minus x, x to the fourth, and x cubed are defined everywhere, then the only critical points are going to be the solutions to derivative equal to zero. Remember the reason why factoring is useful is that if a product of two things is zero, we know one of them has to be zero. But in this case, we know that e to the minus x can never be zero. So the only critical points are going to be the solutions to minus x to the fourth plus four x cubed equals zero. So again, we'll try to factor by identifying a common factor to all the terms and we can remove a factor of x cubed. And again, we have product equal to zero, so we know that one of the two factors has to be zero. And so that gives us critical points at x equals zero and x equals four. To find the inflection points in the concavity, we'll take the second derivative. And here's where that partial factorization is useful because now our first derivative is a product and so we can apply the product rule. And again, looking ahead, it doesn't hurt to do some partial factorization here. Both of these terms have a factor of e to the minus x that we can remove and we get our partially factored form of the second derivative. Now we want to find the inflection points and that's going to be where our concavity changes. But remember that if our second derivative is positive, the graph is concave up. If the second derivative is negative, the graph is concave down. What that means is that the inflection points must be where our second derivative is either zero or undefined. And once again, e to the minus x and polynomials are defined everywhere so our second derivative will never be undefined. So let's look for where it is equal to zero. So we've found the second derivative as a product of two things. So if we want it to be equal to zero, we need one of the factors to be zero. Either e to the minus x is zero, which is impossible, or the other factor is zero. And again, let's factor. And now I have a product of two things equal to zero. So either x equals zero or the second factor equals zero. Now this is a quadratic equation and in a kind and gentle universe, every quadratic equation has integer solutions. We do not live in that universe so we'll use the quadratic formula. Okay, so every now and then we get lucky and we do have a quadratic with integer solutions. And so we have our derivatives in factored form. We have the places where the first derivative is zero. We have the places where the second derivative is zero. So now let's fill in the signs at all the other locations. If x is less than zero, then in the derivative, e to the minus x is positive. And minus x to the fourth plus four x cubed is negative. So the derivative will be the product of a positive and a negative. It'll be negative. And we'll make a note of that on our sign chart. Meanwhile, if x is less than zero, our second derivative will be a product of a positive number and a positive number. So the second derivative will be positive. Similarly, if we're between zero and two, e to the minus x is positive and minus x to the fourth plus four x cubed is also going to be positive. So the derivative is positive. And in this way we can find the sign of the derivative and the second derivative in between the critical points. Now based on this, we see that the graph is falling until we reach x equals zero and rising after. So a local minimum occurs at x equals zero. The graph continues to rise until x equals four and then falls afterward. So a local maximum occurs at x equals four. Likewise, we see that the graph is concave up until x equals zero and concave up afterwards. So no changing concavity occurs at x equals zero. But it does switch to being concave down at x equals two and then switches back to being concave up at x equals six. So we have an inflection point at x equals two and x equals six. So all of the interesting behavior of this graph is going to occur at x equals zero, x equals four, x equals two and x equals six. So any interval that includes zero, four, two and six will be good for graphing. So a suitable interval for graphing might be the interval between minus one and seven. And if we do that, our graph looks like this.