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y=ln([2x-1/2x=1]^1/2)- can be written as y= [0.5ln(2x-1)]-[0.5ln(2x+1)] due to laws of logs. Take first term -- (0.5ln(2x-1)) and substitute 2x-1 for u. so u=2x-1 and y=0.5lnu Now dy/du=1/2u and du/dx=2. ...

In order to answer this question we will break it down into several pieces. Firstly, using physical arguments, can we narrow down which interval the man must overtake the women? After some thought, it mus...

Generally, quadratic equation of the form $\color{orange}{a}x^2 - \color{blue}{b}x + \color{green}{c} = 0$ where $\color{orange}{a} \neq 0$ can be solve by evaluating $$\color{brown}{\Delta} = \color{blue...

At first this equation seems tricky, but we can perform a clever substitution to simplify it. We notice that if let y = 2^x, then we can rewrite this as:

8(y^2) - 9y + 1 = 0

This now becomes...

A stationary point is where the gradient is exactly zero - the curve is neither increasing or decreasing. This means that we need to differentiate y to find dy/dx and then set this equal to 0. Doing this,...