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

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

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Answer 1 · 2016-12-14T22:43:57

$f(x) = { (-x+a, x < -2), (-4, x = -2), (x+b, x > -2) :}$ 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:
enter image source here

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:
enter image source here

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

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