a differentiable function y =f(x) that has the given properties.

1. local minimum at (1,1), and local maximum at (3,3)

2. local minima at (1,1) and (3,3)

3. local maxima at (1,1) and (3,3)

I don't remember how to do this. Thank you.

http://en.wikipedia.org/wiki/Maxima_and_minima

I don't see how any function can have a local minima and maxima at the same point.

That is 3 different problems.

Explain Bot Explain Bot answered
9 months ago

No problem! I can help you with these questions. Let's go through each scenario one by one.

1. To find a function with a local minimum at (1,1) and a local maximum at (3,3), we can start by considering a quadratic function, as it can easily achieve these properties. Let's suppose our function is y = ax^2 + bx + c, where a, b, and c are constants.

To find the values of a, b, and c, we can set up a system of equations based on the conditions given. Since we want a local minimum at (1,1), we know that the derivative of the function at x = 1 should be zero, and the second derivative should be positive. Similarly, for the local maximum at (3,3), the derivative at x = 3 should be zero, and the second derivative should be negative.

Let's differentiate the quadratic function y = ax^2 + bx + c with respect to x to find the derivative:

dy/dx = 2ax + b

To find the second derivative, we differentiate the derivative with respect to x:

d^2y/dx^2 = 2a

Now, let's substitute x = 1 and x = 3 into the derivative and second derivative equations:

At x = 1:
dy/dx = 2a + b
d^2y/dx^2 = 2a

At x = 3:
dy/dx = 6a + b
d^2y/dx^2 = 2a

From the conditions given, we know that dy/dx = 0 at both x = 1 and x = 3. Hence, we have the following equations:

2a + b = 0 (equation 1)
6a + b = 0 (equation 2)

By solving this system of equations, we can find the values of a and b. Once we have a and b, we can substitute them into the original quadratic function to find c, using either of the points (1,1) or (3,3).

2. For a function with local minima at (1,1) and (3,3), we can use a similar approach. However, instead of having opposite signs for the second derivatives at the two points, we can consider the second derivatives to be positive at both points. The rest of the process would be the same as in scenario 1.

3. Similarly, for a function with local maxima at (1,1) and (3,3), we can consider the second derivatives to be negative at both points. The rest of the process remains the same.

Remember, this method is specific to quadratic functions and might not work for all types of functions. For more complex functions, you may need to use optimization techniques or consult mathematical software.

I hope this helps you understand how to approach these problems systematically. If you have any further questions, feel free to ask!