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Solve the inequality x^2 - 9 > 0

This is a quadratic inequality, because we have an x 2 term, so we answer this question by examining the graph of the associated equation y = x 2 - 9, and then find out where this graph is greater than 0.So ...
RH
Answered by Rose H. • Maths tutor
18268 Views

Find the x coordinate of the stationary points of the curve with equation y = 2x^3 - 0.5x^2 - 2x + 4

Firstly, to find the stationary points of a curve you must differentiate the equation of the curve. To do this each x component is multiplied by its current power and then the power is decreased by one. Any ...
BS
Answered by Bartosz S. • Maths tutor
6442 Views

Use the substitution u = cos 2x to find ∫(cos^2*(2x) *sin3 (2x)) dx

∫(cos 2 2x *sin 3 2x)dx u = cos2x - u =(du/dx) = -2sin2x - differentiate u dx = du/(-2sin(2x)) - dx = -1/2 ∫cos 2 2x * sin 2 2x du - sub in dx-1/2 ∫u 2 (1-u 2 )du - put in terms if u -1/2 [ u 3 /3 - u 5 /5 ]...
WB
Answered by Will B. • Maths tutor
8305 Views

Does the equation x^2 + 2x + 5 = 0 have any real roots?

This equation has no real roots, as when we test the equation with the formula b 2 - 4ac, we get 2 2 -4 1 5 = 4 - 20 = -16. As this is negative we can be assured that this equation has no real roots.
SB
Answered by Sami B. • Maths tutor
5878 Views

Find the gradient of the tangent and the normal to the curve f(x)= 4x^3 - 7x - 10 at the point (2, 8)

y = 4x 3 - 7x -10The gradient of the function at any point can be found using its derivative:dy/dx = 12x 2 - 7The gradient of the function, m 1 , at (2,8) is equal to the gradient of the tangent at that poin...
MP
Answered by Miss P. • Maths tutor
5998 Views