How do you find the value of c such that #y=3/2x+6# is tangent to the curve #y=csqrtx#?

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Answer 1 · 2018-03-07T00:34:49

#c= 6#

we start by finding the derivative of the curve.

#y' = c/(2x^(1/2))#

We know that the value of the tangent is the slope of the tangent line. Thus

#3/2 = c/(2x^(1/2))#

#c = 3x^(1/2)#

So now we have a system of three equations with the same number of unknowns. That system is #{(c= 3x^(1/2)),(y = csqrtx), (y = 3/2x + 6):}#

If we substitute the first and third equations into the second, we get:

#3x^(1/2)sqrt(x) = 3/2x + 6#

#3x = 3/2x + 6#

#3/2x =6#

#x= 4#

Therefore we get that #c =3(4)^(1/2) = 3(2) = 6#. A graph confirms:

enter image source here

Hopefully this helps!