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Find the turning points on the curve with the equation y=x^4-12x^2

y = x^4 - 12x^2
dy/dx = 4x^3 - 24x
The turning points are where dy/dx = 0
4x^3 - 24x =0
x(4x^2 - 24) = 0 Therefore one of the turning points is at x = 0
4x^2 - 24 = 0
4x^2 =...

Answered by Maths tutor
4436 Views

Solve 7(x+2)=5x+21

The first thing we must realise is that there are brackets in this equation and thus this has to be dealt with. We expand out the brackets on the left side by multiplying 7 by x and 2 to obtain '7x+14'. W...

Answered by Maths tutor
3206 Views

Draw a graph of x = (y+4)/4.

Note: There are multiple ways to do this, so find a way that works for you!Step 1: Get rid of the fraction.Step 2: Rearrange the new formula into y = mx + c.Step 3: Find the y-intercept and the x-intercep...

TN
Answered by Tarin N. Maths tutor
2700 Views

How do you solve simultaneous equation where one of them involves powers?

Example: y - 3x = 2 , x2 + y2 = 4 y= 3x + 2 x2 + (3x+2)2 = 4 10x2+ 12x + 4 = 4 10x2+12x=0 x(10x+12)=0 x=0 or x=-1.2 y=30 + 2 = 2 (0,2)...

LF
Answered by Luke F. Maths tutor
2615 Views

Show that arctan(x)+e^x+x^3=0 has a unique solution.

Since either sketching the function f(x)=arctan(x)+ex+x3 or evaluating the precise/approximated solutions of the equation would be impossible with A-level techniques, we have to come...

Answered by Maths tutor
3264 Views

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