More Discrminant help simple A-level

hi, i need help on these. These are not the only questions ive been set, ive been set a lot, so please dont feel like your doing my homework. Can you go through all the working out you do. thanks:

1. Find the gradient of the function at the point indictated:
F(x)= 2-3x^2 at the point (-1,5)

2. Find the coordinates of the points on the curve y=x^3+2x-1 where the gradient is 29

3. Given that y= (2x^2-3)^2, show that dy/dx = 8x(2x^2-3)

Many thanks :D

1. y = f(x) = 2 - 3x^2

f '(x) = - 6x

at (-1, 5) f '(x) = - 6 * - 1

= 6

(By the way f '(x) = dy/dx = Gradient)


2. y = x^3 + 2x - 1

dy/dx = 3x^2 + 2

29 = 3x^2 + 2

3x^2 = 27

x^2 = 9

x = + or - 3

y = x^3 + 2x - 1

= 32 or - 34

(3, 32) and (- 3, - 34)

3. y = (2x^2 - 3)^2

u = 2x^2 - 3
Therefore y = u^2
dy/du = 2u

u = 2x^2 - 3

du/dx = 4x

dy/du * du/dx = dy/dx

so dy/dx = 2u * 4x

= 2(2x^2 - 3) * 4x

= 8x(2x^2 - 3)

Hope this helped :)

Hmm.

The gradient is the first derivative.

For the gradient at a point, take the first derivative and plug in the given value of x.

F(x) = 2 - 3x^2

F'(x) = - 6x

F'(- 1) = - (6)(- 1) = 6


y = x^3 + 2x - 1

y' = 29 = 3x^2 + 2

Simplify

27 = 3x^2

9 = x^2

x = 3, x = - 3

Plug those values into the original function to determine the y-coordinates.


y = (2x^2 - 3)^2

Using the power rule,

dy/dx = 2[(2x^2 - 3)](4x) = 8x(2x^2 - 3)