Given $y = \frac{4}{\sqrt{2x-3}}$, $x 1\frac{1}{2}$, find \(\frac{\mathrm{d}y}{\mathrm

Question

Given $y = \frac{4}{\sqrt{2x-3}}$, $x > 1\frac{1}{2}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$. Explain why the graph of $y = \frac{4}{\sqrt{2x-3}}$ does not have a stationary point. State whether $y = \frac{4}{\sqrt{2x-3}}$ is an increasing or decreasing function. Explain your answer clearly.

Accepted Answer

$\frac{\mathrm dy}{\mathrm dx}=-\frac4{(2x-3)^{3/2}}$ (a) No stationary point because $\frac{\mathrm dy}{\mathrm dx}\ne0$ (b) Decreasing for $x>\frac32$

Equations without Stationary Points

Equations without Stationary Points

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