Question

How do you sketch the graph that satisfies f'(x)=1 when x>-2, f'(x)=-1 when x<-2, f(-2)=-4?

Answer 1

`f(x) = { (-x+a, x < -2), (-4, x = -2), (x+b, x > -2) :}` where `a,b` are constants

Explanation:

For `x < -2` we have `f'(x)=-1 => f(x)=-x+a`, ie a straight line with gradient `-1`

For `x > -2` we have `f'(x)=1 => f(x)=x+b`, ie a straight line with gradient `1`

We want `f(-2)=-4` and we are not told anything about the gradient when `x=-2` (and in fact `f'(-2)` will be undefined) and we are not told that that the function should be continuous.

Combining the results we get:

`f(x) = { (-x+a, x < -2), (-4, x = -2), (x+b, x > -2) :}` where `a,b` are constants

If we wanted a continuous solutions then we would need:

For `x < -2 => 2+a = -4 \ \ \ \ => a=-6`
For `x > -2 => -2+b = -4 => b=-2`

And so:

`f(x) = { (-x-6, x < -2), (-4, x = -2), (x-2, x > -2) :}`

Which we can graph as follows:

But equally we could choose any values for the arbitrary constants `a` and `b`, for example we could choose `a=b=0` to get

`f(x) = { (-x, x < -2), (-4, x = -2), (x, x > -2) :}`

Which we can graph as follows: