# Thread: Eqn of tangent line help

1. ## Eqn of tangent line help

I'm having trouble finding a value of x for f(x) whose tangent line has the slope 4/3. I am not given any information regarding points on the graph.

The following is the original function

$f(x)=3x^2/(2x+1)$What I have done so far is found the derivative and set that equal to 4/3 and then solve for x. This has not worked for me so far. What should I do?

2. Originally Posted by cognoscente
I'm having trouble finding a value of x for f(x) whose tangent line has the slope 4/3. I am not given any information regarding points on the graph.

The following is the original function

$f(x)=3x^2/(2x+1)$What I have done so far is found the derivative and set that equal to 4/3 and then solve for x. This has not worked for me so far. What should I do?

f'(x) = [6x(2x+1) - 6x^2]/(2x+1)^2 = 6[x^2 + x]/(2x+1)^2 = 4/3 ==>

[x^2 + x)]/(2x + 1) = 2/9 ==> 9x^2 + 9x = 4x + 2 ==> 9x^2 + 5x - 2 = 0

This quadratic's discriminant is D = 25 + 72 = 97 ==> there are two real solutions:

x_1,2 = [-5 (+/-) Sqrt(97)]/18 , and since both these points are different

from -1/2, where the function isn't defined, then we're done.

Tonio