Question

Find a function whose graph has a relative minimum when x = 1 and a relative maximum when x = 4?

Answer 1

Many answers possible, for example `y = 5/2x^2 - 1/3x^3 - 4x + 6`

Explanation:

Suppose this graph has two critical points, one at `x= 1` and one at `x= 4`. Then by the definition of critical points,

`(x- 1)(x- 4) = 0`

If you test, you will see that at `x = 2`, `(x- 1)(x- 4)` is negative. Therefore `x= 1` is a maximum which we don't want.

This can be fixed by letting

`-(x- 1)(x -4) = 0`

Now at `x = 2`, the expression `-(x -1)(x -4)` is positive. Similarly, at `x=0`, it's negative, ensuring that `x= 1` is a minimum. At `x = 5`, `-(x- 1)(x - 4)` is negative, ensuring that `x= 4` is a maximum. So the derivative of our function will be

`y' = -(x- 1)(x- 4)`

Let's integrate to find our function!

`y' = -(x^2 - x - 4x +4)`

`y' = -(x^2 - 5x + 4)`

`y' = 5x - x^2 - 4`

`y = 5/2x^2 - 1/3x^3 - 4x +C`

Since no y-values are specified for our max/min, `C` can be any real number.

Thus

`y = 5/2x^2 - 1/3x^3 - 4x + sqrt(2) or y = 5/2x^2 - 1/3x^3 - 4x + 127`

Infinite many answers at this point. Here is the graph of one such function , `y = 5/2x^2 - 1/3x^3 - 4x + 6`.

Hopefully this helps!