# stationary point

• May 10th 2006, 10:52 AM
stationary point
hey having a problem with the following qus.

investigate the function f(x) = 2x - 6/(5-3x) for the nature of the stationary points (if any) and the value of the function at these points.

i found dy/dx = 2 + 18(5-3x)^-2 make it equal to zero

can't seem the solve for x?
• May 10th 2006, 11:17 AM
earboth
Quote:

hey having a problem with the following qus.

investigate the function f(x) = 2x - 6/(5-3x) for the nature of the stationary points (if any) and the value of the function at these points.
i found dy/dx = 2 + 18(5-3x)^-2 make it equal to zero
can't seem the solve for x?

Hello,

you've made a minor mistake because you forgot to use the chain rule:

$\frac{dy}{dx} = 2 -6 \cdot (-1) \cdot (5-3x)^{-2} \cdot (-3)=2-\frac{18}{(5-3x)^2}$

Set this term equal to zero.

Multiplying by the denominator and dividing by 2 will give:

$(5-3x)^2=9$

I'll leave this to you.

Greetings

EB
• May 10th 2006, 11:29 AM