Question

Minimum and Maximum points?

The equation of a curve is such that `(d^2y)/dx^2 = 2x - 1`. Given that the curve has a minimum point at `(3, -10)`, find the coordinates of the maximum point.

Answer 1

Please see below.

Explanation:

`(d^2y)/dx^2 =2x-1` and maximum point is `(3,-10)`

Let `f(x) = y`

A local maximum occurs at a critical number, so `f'(3) = 0` or #f'(3) is not defined.

`dy/dx = f'(x) = x^2-x+C`

Clearly `f'(x)` is defined for all `x`, so we can find `C` by using `f'(3) = 0`

`C = -6`

So `dy/dx = f'(x) = x^2-x-6`

Soloving `f'(x) = 0`, we see that there is another critical point at `x = -2`.

Since `f''(-2) < 0` we know that `f(-2)` is a local maximum.

We were asked for the maximum "point" which means we want the two coordinate point on the curve. SO we need to find an expression for `y` (for `f'(x)`).

From `dy/dx = f'(x) = x^2-x-6` we get

`y = f(x) = x^3/3-x^2/2-6x+D`

Use the given point, which tells us that `f(3) = -10`, to find `D = 7/2` and

`y = f(x) = x^3/3-x^2/2-6x+7/2`

At `x = -2`, we get `f(-2) = 65/6`

So the maximum point is `(-2,65/6)`