y = (x^2 + 3)/(2x^2)

=> y = (x^2)/(2x^2) + 3/(2x^2)

=> y = 1/2 + (3/2)x^-2

=> y' = -3x^-3

we want y' = 3

=> 3 = -3x^-3

=> -1 = x^-3

=> -1 = x^3 ..........i took the inverse

=> x = -1

when x = -1, y = [(-1)^2 + 3]/[2(-1)^2] = 4/2 = 2

so the coordinate where the gradient is 3 is (-1,2)

y = Ax^2 + Bx

Question 2:

Given that the curve with equation y=Ax^2 + Bx has gradient 7 at the point (6,8), find the values of the constants A and B.

=> y' = 2Ax + B

when x = 6, y' = 7

=> 7 = 12A + B ..............(1)

also (6,8) is a point on the curve, so going back to the original equation and plugging these in we get:

8 = 36A + 6B

=> 4 = 18A + 3B ..............(2)

so we have the simultaneous equations:

7 = 12A + B ..............(1)

4 = 18A + 3B ............(2)

=> 4/3 = 6A + B .........(3) = (2)/3

=> 17/3 = 6A

=> A = 34

but 7 = 12A + B

=> B = 7 - 12(34) = -401

so A = 34, B = -401 ..........wierd numbers I know

when y = 9Question 3:

The two tangents to the curve y=x^2 at the points where y = 9 intersect at the point P. Find the coordinates of P.

=> 9 = x^2

=> x = +/- 3

so the tangents cut the curve at (3,9) and (-3,9). what are the equations of these tangent lines? why are we interested in that. if we find the equations of the tangent lines, we can equate them and find where they intersect. we can find the gradient from the derivative

y' = 2x

so at (3,9), gradient is 2(3) = 6

at (-3,9), gradient is 2(-3) = -6

for the line passing through (3,9)

y - y1 = m(x - x1)

=> y - 9 = 6(x - 3) = 6x - 18

=> y = 6x - 9

for the line passing through (-3,9)

y - y1 = m(x - x1)

=> y - 9 = -6(x + 3) = -6x - 18

=> y = -6x - 9

so we have y = 6x - 9 and y = -6x - 9

so they intersect when 6x - 9 = -6x - 9

=> 12x = 0

=> x = 0

so they intersect at (0,-9), which is the point P

the gradient is given by the derivativeQuestion 4:

Find the values of x at which the gradient of the curve

y = x^3 - 6x^2 + 9x + 2 = 0

y = x^3 - 6x^2 + 9x + 2

=> y' = 3x^2 - 12x + 9

set y'=0

=> 0 = 3x^2 - 12x + 9

now we solve for x

=> 0 = x^2 - 4x + 3

=> (x - 3)(x - 1) = 0

so x = 3, x = 1

these are the x-values for which the condition holds

Disclaimer: the methods are correct, but you might want to double check my computations, i've been making stupid mistakes all week