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March 20th, 2016, 02:30 AM   #1
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How to find the following tangent line


I am given f(x)=3sqrt(x).
So I know that f'(x)=(3/(2sqrt(x))),

I need to find the tangent to f(x) that goes through (-3,-3)..

I know that y=mx+n, and I have a point and the tangent, but I seem to be missing a detail to solve it.

Any ideas?

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March 20th, 2016, 08:41 AM   #2
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Math Focus: tetration
""> y = 1.5x + 1.5
Thanks from noobinmath

Last edited by manus; March 20th, 2016 at 09:31 AM.
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March 20th, 2016, 01:06 PM   #3
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Thanks, this is the correct answer, but what is the way?
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March 20th, 2016, 01:49 PM   #4
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Let $(x_0,3\sqrt{x_0})$ be the point where the tangent line touches the graph. The gradient of this line is then given by $f'(x_0) = \dfrac{3}{2\sqrt{x_0}}$.

However, the gradient is also given by $\dfrac{3\sqrt{x_0} + 3}{x_0 + 3}$ (can you think why?).

Now you have the equation $\dfrac{3}{2\sqrt{x_0}} = \dfrac{3\sqrt{x_0} + 3}{x_0 + 3}$ solve for $x_0$, then find the gradient. After that, you have all the information you need.
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March 23rd, 2016, 10:31 AM   #5
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The function is $\displaystyle f(x)= 3\sqrt{x}$ and the derivative is $\displaystyle f'(x)= \frac{3}{2\sqrt{x}}$ as said. So any tangent to the curve at $\displaystyle \left(x_0, 3\sqrt{x_0}\right)$ is of the form $\displaystyle y= \frac{3}{2\sqrt{x_0}}\left(x- x_0\right)+ 3\sqrt{x_0}$. Since that tangent line must go through (-3, -3), we must have $\displaystyle \frac{3}{2\sqrt{x_0}}\left(-3- x_0\right)+ 3\sqrt{x_0}= -3$. Solve that equation for $\displaystyle x_0$.
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