Find a and b such that y=asrtx+ b/srtx has (1,8) as a point of inflection.

Find a and b such that y=a sqrt x + b/sqrtx has (1,8) as a point of inflection.

y = a√x + b/√x

y' = a/2√x - b ÷ 2x^3/2 = (ax-b) ÷ 2x^3/2
y'' = (3b-ax) ÷ 4x^5/2

so, y''=0 when x = 3b/a

3b/a = 1 ==> a = 3b
y(1) = 3b + b = 4b = 8 ==> b=2, so a=6

y = 6√x + 2/√x

To find the values of a and b such that the equation y = asqrt(x) + b/sqrt(x) has (1,8) as a point of inflection, we need to determine the conditions for a point of inflection.

A point of inflection occurs when the second derivative of the function changes signs. Hence, we need to find the second derivative of the given equation and set it equal to zero to find the critical points.

Let's start by finding the first derivative of y with respect to x:

dy/dx = d/dx (asqrt(x) + b/sqrt(x))
= a / (2sqrt(x)) - b / (2x^(3/2))

Now, to find the second derivative, we differentiate dy/dx:

d²y/dx² = d/dx (a / (2sqrt(x)) - b / (2x^(3/2)))
= -a / (4x^(3/2)) + 3b / (4x^(5/2))

We set d²y/dx² equal to zero to find the x-coordinate of the point(s) of inflection:

-a / (4x^(3/2)) + 3b / (4x^(5/2)) = 0

To make it easier, we can multiply through by 4x^(5/2):

-a * x + 3b = 0

Now, since (1,8) is a point of inflection, the x-coordinate must satisfy this equation. So, we substitute x = 1 into the equation and solve for a and b:

-a(1) + 3b = 0
- a + 3b = 0
3b = a

Given that the point (1,8) is also on the curve, we substitute x = 1 and y = 8 into the equation and solve for a and b:

8 = a * sqrt(1) + b / sqrt(1)
8 = a + b

Now, we have a system of equations:

3b = a
8 = a + b

Solving this system of equations, we find:

a = 4
b = 4

Therefore, a = 4 and b = 4 are the values that satisfy the condition for the point (1,8) to be a point of inflection in the equation y = asqrt(x) + b/sqrt(x).