# Find the value of c such that the line y=1/6x+3/2 is tangent to the curve y=(c)sqrt(x) Tangent means the functions pass through the same point and have the same slope at that point.

So the derivatives must be equal:

`1/6=c/(2sqrt(x))`

And the y-values must also be equal (since the graphs pass through the same point):

`1/6x+3/2=csqrt(x)`

So we have a system of two equations in two...

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Tangent means the functions pass through the same point and have the same slope at that point.

So the derivatives must be equal:

`1/6=c/(2sqrt(x))`

And the y-values must also be equal (since the graphs pass through the same point):

`1/6x+3/2=csqrt(x)`

So we have a system of two equations in two variables.  Let's solve each for c and then set them equal.

`sqrt(x)/3=c`

`sqrt(x)/6+3/(2sqrt(x))=c`

Setting these equal we have

`sqrt(x)/3=sqrt(x)/6+3/(2sqrt(x))`

`sqrt(x)/6=3/(2sqrt(x))`

Cross-multiplying

`2x=18`

Cha-ching!

x = 9 so c = 1.

Let's double-check that x/6+3/2 and sqrt(x) are tangent at the point (9,3):

Yeah.

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