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Find the x-coordinates of any stationary points of the equation y = x^3 - 2x + 4/x

y = x^3 - 2x +4/x, dy/dx = 3x^2 - 2 - 4/(x^2) = 0 at the stationary points, 3x^4 - 2x^2 - 4 = 0, substitute in u for x^2: 3u^2 - 2u - 4 = 0, use the quadratic formula: u = (-(-2) +- sqrt((-2)^2 - 4

Answered by • Maths tutor

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60 students were taking a Maths, Physics or Chemistry exam. 38 of the students were male. 11 of the 32 students who were taking the Maths exam were female. 8 males were taking the Physics exam. 12 students were taking the Chemistry exam. One of the fe

60-38 = 22 females 32+12=44 students taking the Chemistry and Maths exam. Therefore 60-44=16 students were doing the Physics exam, so 16-8= 8 females were taking the Physics exam. This means that 22-11-8 ...

Answered by Harry B. • Maths tutor

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What is the relationship between the lengths of a right angle triangle

The lengths of a right angle triangle can be summarised as follows: the sum of the lengths of the two shorter sides squared (the sides which vertice connect forms the right angle) is equal to square of th...

Answered by stephanie g. • Maths tutor

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3y^2 + 10y + 3

3y²+ 10y + 33y² = 3 * 1(3y + n) * (y + r)n * r = 3 hence to try whole numbers first n = 1 or 3 meaning r = 3 or 1 respectivey * n + 3y * r = 10yif n =1 and r = 3 then equivalent is ...

Answered by Jack h. • Maths tutor

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Curve C has equation 4x^2- y^3 - 4xy +2^y = 0 , point P (-2, 4) lies on C, find dy/dx at the point P

Use implicit differentiation 1) 8x - 3y^2dy/dx - 4y - 4xdy/dx +2^y*ln2 * dy/dx = 0 You then sub in the points P (-2,4) 2) 8(-2) - 3(4)^2 *dy/dx - 4(4) - 4(-2) *dy/dx + 2^(4) *ln2 * dy/dx = 0 Rear...

Answered by Fernando F. • Maths tutor

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