Thread: Help finding the tangent function??

Let f(x) = (3x^2)/(2x+1). For what values of x does f(x) have a tangent with slope 4/3 ?

Originally Posted by yess

Let f(x) = (3x^2)/(2x+1). For what values of x does f(x) have a tangent with slope 4/3 ?

Slope of the tangent = gradient of tangent=gradient of curve = dy / dx = 4/3

Find dy / dx, for y = f(x), and then form the equation: dy / dx=4/3.
Solve for x.

Originally Posted by Ithaka

Slope of the tangent = gradient of tangent=gradient of curve = dy / dx = 4/3

Find dy / dx, for y = f(x), and then form the equation: dy / dx=4/3.
Solve for x.

aaah i keep getting stuck! i get this big long equation and i dont know how to simplify it! i got:
lim of Δx->0 [(6x²Δx + 6xΔx² + 6xΔx + 3Δx²)/(4x² + 8xΔx + 4x + 1)]/Δx

You don't need to find the derivative by first principles. Just use the product rule.

Or even the quotient rule!

Originally Posted by yess

aaah i keep getting stuck! i get this big long equation and i dont know how to simplify it! i got:
lim of Δx->0 [(6x²Δx + 6xΔx² + 6xΔx + 3Δx²)/(4x² + 8xΔx + 4x + 1)]/Δx

This is the same as


Since there is a " " in each term in the numerator, you can cancel that " " multiplying the denominator:


Now, neither numerator nor denominator goes to 0 as goes to 0.

Originally Posted by HallsofIvy

This is the same as


Since there is a " " in each term in the numerator, you can cancel that " " multiplying the denominator:


Now, neither numerator nor denominator goes to 0 as goes to 0.

so then do i sub in 0 for wherever there is a Δ then equate that to 4/3? because when i factor that out i get x= -2,1 but the question asks for THE x value as in only one??