Creating a Function-Derivatives

[salemchic05]

I need to create a function that has one turning point and one point of inflection. I have no idea what I'm doing!

 

[mathman]

Try xe^-x. The general idea is that is has a max (the turning point) and then asymptotically goes to 0 (as x becomes infinite), so it needs to switch from concave to convex along the way.

 

[M98Ranger]

Creating a function with a turning point and a point of inflection can seem daunting at first, but with some guidance, it can become much easier. A turning point is a point on the graph of a function where the slope changes from positive to negative or vice versa. A point of inflection is where the concavity of the graph changes from concave up to concave down or vice versa.

To create a function with these characteristics, we can start by thinking about the basic shape of the graph we want to create. In this case, we want a graph that has a single hump or bump, with the turning point at the top of the hump and the point of inflection at the bottom. One way to achieve this is by using a quadratic function, which has the general form of f(x) = ax^2 + bx + c.

To ensure that our function has a turning point and a point of inflection, we need to make sure that the coefficient of the x^2 term is not zero. This will give us a parabola that has a single turning point. Next, we need to determine the values of a, b, and c that will give us the desired shape of the graph.

To create a turning point at the top of the hump, we want the coefficient of the x term (b) to be zero. This will eliminate the linear term in our function, creating a symmetrical parabola. Now, we can focus on the value of a, which will determine the steepness of the hump. To create a point of inflection at the bottom of the hump, we want the value of a to be negative. This will give us a parabola that starts off concave up, then becomes concave down, creating the desired point of inflection.

For example, let's say we want our turning point to be at (2, 5) and our point of inflection to be at (4, 3). We can use the information we have to create a system of equations to solve for the values of a, b, and c.

f(2) = 5: 4a + 2b + c = 5
f'(2) = 0: 8a + b = 0
f(4) = 3: 16a + 4b + c = 3
f''(4) = 0: