Question #51aa2

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Answer 1 · 2016-09-21T16:10:35

Please see below.

$y = abs(x^2-1) = {(x^2-1," if ",x^2-1 > 0),(1-x^2," if ", x^2-1 < 0):}$

Solving the inequality $x^2-1 > 0$ gets us,

$y = {(x^2-1," if ",x < -1),(1-x^2," if ",-1 < x < 1),(x^2-1," if ",x > 1) :}$

So the derivative is given by

$y' = {(2x," if ",x < -1),(-2x," if ",-1 < x < 1),(2x," if ",x > 1) :}$

$2x=1$ at $x=1/2$ but the derivative at $x=1/2$ is not given by $2x$ .

$-2x = 1$ at $x= -1/2$ and at $x=-1/2$ , the derivative is given by $-2x$ .

So the slope is $1$ and $x=-1/2$ . At that point, we get $y = 3/4$ .

So the only point on the curve at which the slope is $1$ is $(-1/2,3/4)$ .