Find a function of the form f(x)=a+bsqrtx that is tangent to the line 2y-3x=5 at the point (1, 4)
from 2y-3x=5 , we know that the slope of the tangent must be 3/2
f(x) = a + b√x = a +b(x)^(1/2)
f ' (x) = (1/2)b x^(-1/2) or b/2√x
so b/2√x = 3/2
2b = 6√x
b = 3√x
also (1,4) lies on f(x) = a + bx
4 = a + b
since b = 3√x = 3√1 = 3
then a = 1
f(x) = 1 + 3√x is our function
check:
f ' (x) = (3/2)x^(-1/2)
at (1,4) , slope = (3/2)(1/√1) = 3/2
equation of tangent:
y-4 = (3/2)(x-1)
2y - 8 = 3x - 3
2y - 3x = 5, as given