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Maths
A Level

Find the minimum and maximum points of the graph y = x^3 - 4x^2 + 4x +3 in the range 0<=x <= 5.

First, use the standard method of setting the derivative to be equal to 0 to find the stationary points. This yields the equation 3x^2 - 8x + 4 = 0 and so the stationary points are at x = 2/3 and 2 respec...

Answered by Guy M. Maths tutor
5166 Views

A curve has equation x = (y+5)ln(2y-7); (i) Find dx/dy in terms of y; (ii) Find the gradient of the curve where it crosses the y-axis.

(i) To find the derivative we will use the product rule. Let u = y+5 and v=ln(2y-7). Then dx/dy = du/dyv + udv/dy = ln(2y-7) + (y+5)*2/2y-7 (used the chain rule in 2nd term - can explain this on ...

Answered by Szymon P. Maths tutor
9725 Views

Solve the equation: log5 (4x+3)−log5 (x−1)=2.

As both terms on the left hand side have base 5 we know we can combine them. When dealing with logs, a minus means we can divide them, and a plus means we can multiply them. This will leave us with log5(4...

Answered by Hugh G. Maths tutor
8239 Views

Show that 1+cot^2(x)=cosec^2(x)

Need to remember that:sin2(x)+cos2(x)=1 (eq.1)Divide the whole eq.1 by sin2(x)to get:sin2(x)/sin2x+cos2(x)/sin2(x)=1 /sin2...

Answered by Tutor114151 D. Maths tutor
49873 Views

Find the stationary point(s) of the curve: y = 3x^4 - 8x^3 - 3.

Firstly. Recognise which method you should use to approach this question. In this case, you can find the stationary point of a curve where its gradient is 0 i.e. at a point where the grad...

Answered by Laurene L. Maths tutor
4461 Views

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