anonymous
  • anonymous
Find the slope of the curve at the indicated point. f(x)=|x| a.) x=2 b.) x=-3
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
For \(x>0\), you have \(|x|=x\). For \(x<0\), you have \(|x|=-x\). So at x=2, \(f'(x)=\dfrac{d}{dx}[x]\bigg|_{x=2}\) And at x=-3, \(f'(x)=\dfrac{d}{dx}[-x]\bigg|_{x=-3}\)
anonymous
  • anonymous
what is the d/dx I don't think i have learned about that
anonymous
  • anonymous
That's the derivative operator. \(f'(x)=\dfrac{d}{dx}[f(x)]\). Is this for calculus?

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anonymous
  • anonymous
yea but i barely started and havent got to that point
anonymous
  • anonymous
Well it's the exact same thing as what you already (presumably) have been taught. If you're more comfortable with \(f'\) notation, I'll stick to that. Since \(f'(x)\) gives you the slope of the tangent line to \(f(x)\) at some value of \(x\), all you have to do is find the derivative of \(f(x)\), then plug in the desired value. So, like I mentioned before, you have \[f(x)=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\] The first value you're given is 2, which is positive, so you use the first piece of the function; \(f(x)=x\). Find \(f'(x)\), then plug in 2. In other words, find \(f'(2)\).
anonymous
  • anonymous
oh so that's is it? Thank you!
anonymous
  • anonymous
Yep. Just keep in mind that part (b) has \(f(x)=-x\).

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