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A curve has equation y = 20x −x^2 −2x^3 . The curve has a stationary point at the point M where x = −2. Find the x-coordinate of the other stationary point of the curve.

dy/dx = 20 - 2x - 6x 2 = -6x 2 - 2x + 20 which factorises to dy/dx = (10 - 6x)(2 + x). This shows that x=-2 and x=10/6=5/3 are stationary points of y = 20x −x 2 −2x 3 .
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Answered by Xixi Y. • Maths tutor
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Find all values of x in the interval 0 ≤ x ≤ 2pi for 2sin(x)tan(x)=3

This questions involves the use of trigonometric identities, specifically [1] tan(x) = sin(x)/cos(x) and [2] sin 2 (x) + cos 2 (x) = 1. Start by rearranging the given equations to tan(x) = 3/2sin(x) by divin...
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Answered by Jasmin C. • Maths tutor
7682 Views

Find the derivative of sin(x)/x^3 with respect to x

First, bring the x to the numerator (top) as x^(-3). Then use the chain rule: State the first times derivative of the second plus state the second times the derivative of the first.State sin(x), then multipl...
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Answered by Tutor170145 D. • Maths tutor
7788 Views

Which value of x gives the greatest value of "-x^2+8x-6"

There are 2 methods of solving this: completing the square or calculus, which makes this a versatile question for both GCSE and A Level students. It is also fairly simple to adapt. Using calculus, the equati...
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Answered by Tutor172970 D. • Maths tutor
3904 Views

y =(4x)/(x^2+5) (a) Find dy/dx, writing your answer as a single fraction in its simplest form. (b) Hence find the set of values of x for which dy/dx<0

dy/dx = 4/(x^2+5) - 4x(2x)/(x^2+5)dy/dx= 4(5-x^2)/(x^2+5)^2Part B4(5-x^2)/(x^2+5)^2&lt;05-x^2&lt;05&lt;x^2
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Answered by Dominic K. • Maths tutor
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