# Thread: basic calc two problems! thanks for the help!

1. ## basic calc two problems! thanks for the help!

ok two problems both are totally weird and have no context in my book

1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)

any help is appreciated
thank you for all your help!

ps
this is my first chapter using calc so like limits and derivatives ect.

2. Originally Posted by wpski
...

1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

...
Hello,

you have:

$\displaystyle f(x)=\frac{ax+b}{x^2} = \frac ax + \frac b{x^2}$

and the equation of a tangent line:

$\displaystyle t:\ y=2x-1$

1. Calculate the coordinates of the tangent point for x = 1. Plug in this value into the equation of the tangent line: T(1, 1)
Plug in these values into the equation of the function because the coordinates of the tangent point must satisfy the equation of the function too:

$\displaystyle f(1) = 1 =\frac{a+b}{1}~\implies~\boxed{a+b=1}~\color{blue}[1]$

2. The slope of the tangent line is m = 2 and is equal to the first derivatie of f at x = 1. Calculate the first derivatie. I took the 2nd version of the term of the function:

$\displaystyle f'(x) = -\frac a{x^2} - \frac{2b}{x^3}$ . Thus

$\displaystyle f'(1) = -\frac a{1} - \frac{2b}{1}~\implies~\boxed{-a-2b = 2}~ \color{blue}[2]$

3. Solve the system of simultaneous equations for a and b. I've got a = 4 and b = -3

3. Originally Posted by wpski
...

2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)

...
Hi,

you have: $\displaystyle f(x) = -\frac{2}{(ax+1)^2} = -2 \cdot (ax+1)^{-2}$

The point T(0, -2) belongs to the graph of f and is the tangent point of a line which passes through the points P(1, 0) and T(0, -2).

Use the 2-point-formula to calculate the equation of the tangent line:

$\displaystyle \frac{y-0}{x-1} = \frac{-2-0}{0-1} = 2~\implies~y = 2x-2$

Thus the slope of the tangent line is m = 2

The slope of the tangent line is equal to f'(0)

$\displaystyle f'(x) = (-2) \cdot (-2) \cdot (ax+1)^{-3} \cdot a = \frac{4a}{(ax+1)^3}$ . That means $\displaystyle f'(0) = \frac{4a}{1^3}=4a$

So you get:

$\displaystyle 4a = 2~\iff~a=\frac12$