A curve has equation y-2x-1-1-2x- x-12Find the x-coordinates of the stationary points

y=(2x 1)^ 1+2x dydx= ddx(2x 1)^ 1+2x dydx= 2(2x 1)^ 2+2 For stationary points #160; dydx=0 2(2x 1)^ 2+2=0 (2x 1)^ 2=1 2x 1=1 ...

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Standard XII

Mathematics

Stationary Point

A curve has e...

Question

A curve has equation $y = (2 x - 1)^{- 1} + 2 x$ . $x \neq \frac{1}{2}$
Find the x-coordinates of the stationary points

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Solution

$y = (2 x - 1)^{- 1} + 2 x$

$\frac{d y}{d x} = \frac{d}{d x} (2 x - 1)^{- 1} + 2 x$

$\frac{d y}{d x} = - 2 (2 x - 1)^{- 2} + 2$

For stationary points
$\frac{d y}{d x} = 0 - 2 (2 x - 1)^{- 2} + 2 = 0 (2 x - 1)^{- 2} = 1 2 x - 1 = 1 x = 1$

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A curve has equation y=(2x−1)−1+2x. x≠12Find the x-coordinates of the stationary points

y=(2x−1)−1+2xdydx=ddx(2x−1)−1+2xdydx=−2(2x−1)−2+2For stationary points dydx=0−2(2x−1)−2+2=0(2x−1)−2=12x−1=1x=1

A curve has equation $y = (2 x - 1)^{- 1} + 2 x$ . $x \neq \frac{1}{2}$
Find the x-coordinates of the stationary points

$y = (2 x - 1)^{- 1} + 2 x$

$\frac{d y}{d x} = \frac{d}{d x} (2 x - 1)^{- 1} + 2 x$

$\frac{d y}{d x} = - 2 (2 x - 1)^{- 2} + 2$

For stationary points
$\frac{d y}{d x} = 0 - 2 (2 x - 1)^{- 2} + 2 = 0 (2 x - 1)^{- 2} = 1 2 x - 1 = 1 x = 1$