# Thread: Help finding the tangent function??

1. ## 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 ?

2. 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.

3. 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

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

5. Or even the quotient rule!

6. 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
$\lim_{\Delta x\to 0}\frac{6x^2\Delta x+ 6x (\Delta x)^2+ 6x\Delta x+ 3(\Delta x)^3}{(4x^2+ 8x\Delta x+ 4x+ 1)\Delta x}$

Since there is a " $\Delta x$" in each term in the numerator, you can cancel that " $\Delta x$" multiplying the denominator:
$\lim_{\Delta x\to 0}\frac{6x^2+ 6x\Delta x+ 6x+ 3(\Delta x)^3}{4x^2+ 8x\Delta x+ 4x+ 1}$

Now, neither numerator nor denominator goes to 0 as $\Delta x$ goes to 0.

7. Originally Posted by HallsofIvy
This is the same as
$\lim_{\Delta x\to 0}\frac{6x^2\Delta x+ 6x (/Delta x)^2+ 6x\Delta x+ 3(\Delta x)^3}{(4x^2+ 8x\Delta x+ 4x+ 1)\Delta x}$

Since there is a " $\Delta x$" in each term in the numerator, you can cancel that " $\Delta x$" multiplying the denominator:
$\lim_{\Delta x\to 0}\frac{6x^2+ 6x\Delta x+ 6x+ 3(\Delta x)^3}{4x^2+ 8x\Delta x+ 4x+ 1}$

Now, neither numerator nor denominator goes to 0 as $\Delta x$ 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??