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f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.

(a) Let y=f(x). Then y = (2x+1)/(x-1). Rearrange the equation to get x in terms of y to obtain the inverse function. This gives x=(1+y)/(y-2). So the inverse of f is f^-1(x)=(1+x)/(x-2)<...

Answered by Lutfha A. • Maths tutor

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Find dy/dx for y = x^3*e^x*cos(x)
In this problem, we see that y is a product of 3 functions of x. That means that in order to find dy/dx we need to use the product rule. The product rule tells us that in this case we should differentiate...
Answered by Lyudmil N. • Maths tutor
9397 Views

Solve x² ≥ | 5x - 6 | (Question from AQA Core 3 June 2016)
To solve this question we can break it down in to a multiple small stages.

The first thing to tackle is the use of modulus around 5x - 6. This effectively means we have two equations to solve:
Answered by James H. • Maths tutor
6041 Views
Use chain rule and implicit differentiation to find dy/dx for y^3 = 1 + 3*x^2, then show that they are equal
When using the Chain Rule, it is best to put our equal in a form where there are no operations performed on the subject. i.e. y = (1 + 3*x²)^1/3

Then to help make it clear what...
Answered by Joshua P. • Maths tutor
8206 Views
How do I simplify (1 / [1 + cos(x) ] ) + (1 / [1 - cos(x) ] )?
In order to add the fractions together, we must have a common denominator of the fractions. The simplest way to do this is to make the denominators of the equations the product of the original two denomni...
Answered by Simon C. • Maths tutor
12397 Views
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f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.

Find dy/dx for y = x^3e^xcos(x)

Solve x² ≥ | 5x - 6 | (Question from AQA Core 3 June 2016)

Use chain rule and implicit differentiation to find dy/dx for y^3 = 1 + 3*x^2, then show that they are equal

How do I simplify (1 / [1 + cos(x) ] ) + (1 / [1 - cos(x) ] )?

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