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Find the coordinates of the point on the curve 
y=(2(x-5))/(sqrt(x+1)) where the gradient is 
(5)/(4).

Find the coordinates of the point on the curve \newliney=2(x5)x+1y=\frac{2(x-5)}{\sqrt{x+1}} where the gradient is \newline54\frac{5}{4}.

Full solution

Q. Find the coordinates of the point on the curve \newliney=2(x5)x+1y=\frac{2(x-5)}{\sqrt{x+1}} where the gradient is \newline54\frac{5}{4}.
  1. Find Derivative: First, we need to find the derivative of yy with respect to xx to determine the gradient of the curve. The function is y=2(x5)x+1y = \frac{2(x-5)}{\sqrt{x+1}}. Using the quotient rule, the derivative yy' is given by: y=[x+12][2(x5)(12)(x+1)12]x+1y' = \frac{[\sqrt{x+1} \cdot 2] - [2(x-5) \cdot (\frac{1}{2})(x+1)^{-\frac{1}{2}}]}{x+1} = 2x+1x5x+1x+1\frac{2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}}{x+1}
  2. Set Equal to 54\frac{5}{4}: Next, set the derivative equal to 54\frac{5}{4} to find the x-coordinate where the gradient is 54\frac{5}{4}.
    2x+1x5x+1x+1=54\frac{2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}}{x+1} = \frac{5}{4}
    Cross-multiplying to clear the fraction:
    4(2x+1x5x+1)=5(x+1)4(2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}) = 5(x+1)
    8x+14(x5)x+1=5x+58\sqrt{x+1} - \frac{4(x-5)}{\sqrt{x+1}} = 5x + 5
  3. Simplify and Solve: Simplify and solve for xx:8x+14(x5)x+1=5x+58\sqrt{x+1} - \frac{4(x-5)}{\sqrt{x+1}} = 5x + 5Let's multiply through by x+1\sqrt{x+1} to clear the square root:8(x+1)4(x5)=5x+5x+18(x+1) - 4(x-5) = 5x + 5 \sqrt{x+1}8x+84x+20=5x+5x+18x + 8 - 4x + 20 = 5x + 5 \sqrt{x+1}4x+28=5x+5x+14x + 28 = 5x + 5 \sqrt{x+1}

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