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$f(x) = \sqrt{2x}$

The line tangent to the point $(x_0 , y_0)$ is $y = x + 1/2$

Find $(x_o , y_0)$.

I'm trying to use the point slope formula $$y= f'(x_0) (x-x_0) + y_0$$

But I have nothing to plug in for $x$ and $y$...At least nothing I can spot.

I know $$f'(x) = \frac{1}{\sqrt{2x}}$$

So I could plug that in, but that doesn't seem to clear it up. Some hints appreciated.

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$f'(x_0) = $ the slope of the curve at the point to which the given line is tangent. The slope of $y = x + 1/2$ is $m = 1$.

$$f'(x) = \frac 1{\sqrt{2x}}$$. $f'(x_0) = 1$ means $$f'(x_0) = \frac 1{\sqrt {2x_0}} = 1$$

Now solve for $x_0$: $$\sqrt {2x_0} = 1 \iff 2x_0 = 1 \iff x_0 = \frac 12$$

Use this value $x_0$ and evaluate $f(x_0)$ to find $y_0$.

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  • $\begingroup$ Don't you mean $$f'(x_0) = 1$$? $\endgroup$ – user3200098 May 10 '14 at 14:22
  • $\begingroup$ Great! Thanks. My error was not simply evaluating $f(x_0)$ $\endgroup$ – user3200098 May 10 '14 at 14:24
  • $\begingroup$ Yes, sorry for the typo and its confusion! $\endgroup$ – amWhy May 10 '14 at 14:25
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You wrote your equation wrong. Slope of a line is constant and $f'(x)$ must be $f'(x_0)$

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  • $\begingroup$ I edited but I'm no closer to the answer $\endgroup$ – user3200098 May 10 '14 at 13:43

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