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$y=|x^2-1|$ where $dy/dx=1$

  • $dy/dx=2x$ when $y=x^2-1$
  • $dy/dx=-2x$ when $y=-(x^2-1)$

I managed to find two point after splitting the function into two pieces, but my given answer only accept $(-1/2,3/4)$ while not $(1/2,-3/4)$ any explaination for this? as I do not understand

user317022
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2 Answers2

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The point $(1/2,-3/4)$ is not on the curve, so there is no tangent there. At $x=1/2, y=3/4$ and the slope of the tangent is $-1$

Ross Millikan
  • 374,822
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Note that $$ \left|x^{2}-1\right|=\begin{cases} x^{2}-1 & \text{if }x\geq1\text{ or }x\leq-1;\\ 1-x^{2} & \text{if }x<1\text{ or }x>-1. \end{cases} $$ The derivative is $$ \frac{d}{dx}\left|x^{2}-1\right|=\begin{cases} 2x & \text{if }x>1\text{ or }x<-1;\\ -2x & \text{if }x<1\text{ or }x>-1. \end{cases} $$ Note that the derivative is undefined at $x=\pm1$. Moreover, note that $\pm 2x=1$ at $x=\pm 1/2$. However, only one of these two points is a solution. Can you figure out which one and why? It might help to plot of the derivative.

parsiad
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