4. Calculate the gradient(s) of the curve at the point(s) where it crosses the given line. y = 2x+3/x, y = x
5. Find the coordinates of the point on the curve y = 2x^2 - 5x + 1 at which the gradient it 3.
6. The equation of a curve is y = x + 4/x^2. Find a. the gradient of the curve at the point x = 3. b. the coordinates of the point where the tangent is parallel to the line y + 2 = 0
7. Given that y = 2x^3 - 6x^2 + 4x - 7, find dy/dx and the range of values of x for which dy/dx >= 4.

asked Feb 18, 2013 in CALCULUS

+1 vote

5). y = 2x2 - 5x + 1----------------->(1)

Apply 'derivative with respect to x' each side.

y' = (d/dx)(2x2 - 5x + 1)

y' = (d/dx)(2x2) - (d/dx)(5x) + (d/dx)(1).

Power Rule of Derivative (d/dx)(xn) = nxn-1 and derivative of constant is 0.

y' = 2(2x) - 5(1) + 0

y' = 4x - 5.

But given that gradient it 3. i.e. y' = 3.

3 = 4x - 5.

4x = 8.

Divide each side by 4.

x = 2.

Substitute x = 2 in the equation (1).

y = 2(2)2 - 5(2) + 1

y = 2(4) - 10 + 1 = 8 - 10 + 1 = 9 - 10 = -1.

the coordinates of the point is (x, y) = (2, -1).

+1 vote

y = 2x3 - 6x2 + 4x - 7

Apply 'derivative with respect to x' each side.

(dy/dx) = (d/dx)(2x3 - 6x2 + 4x - 7)

(dy/dx) = (d/dx)(2x3) - (d/dx)(6x2) + (d/dx)(4x) - (d/dx)7.

Power Rule of Derivative (d/dx)(xn) = nxn-1 and derivative of constant is 0.

y' = 2(3x2) - 6(2x) + 4(1) - 0.

y' = 6x2 - 12x + 4.

Given that dy/dx ≥ 4.

So, 6x2 - 12x + 4 ≥ 4

Subtract 4 from each side.

6x2 - 12x ≥ 0.

Take out common term 6x.

6x(x - 2) ≥ 0

6x ≥ 0 or x - 2 ≥ 0

6x ≥ 0 then x ≥ 0.

Take x - 2 ≥ 0

x ≥ 2.

Therefore dy/dx = 6x2 - 12x + 4 and the range of values of x is 0, 2.

4).

The curve is y = (2x + 3)/x and the line is y = x.

To find the gradients of the curve at the points where it crosses the line : y = x, substitute y = x in the given curve.

x = (2x + 3)/x

x2 = 2x + 3

x2 - 2x - 3 = 0

By factor by grouping.

x2 - 3x + x - 3 = 0

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

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

Apply zero product property.

x = 3  and  x = - 1.

Therefore, y = x crosses the curve at x = 3 and x = - 1.

The gradient of the curve y = (2x + 3)/x at any point is given by dy/dx.

Differentiate the curve with respect to x.

dy/dx = (2x + 3)(1/x)' + (1/x)(2x + 3)'

⇒ dy/dx = -(2x + 3)/x2 + (1/x)(2).

When, x = 3, dy/dx = -(2*3 + 3)/32 + (1/3)(2) =  - 1/3.

When, x = - 1, dy/dx = -(2*-1 + 3)/(- 1)2 + (1/(- 1))(2) = 1.

Therefore, gradients of the curve at x = 3 and x = - 1 are - 1/3 and 1.

6).

a).

The curve is y = (x + 4)/x2 and the point is x = 3.

The gradient of the curve y = (x + 4)/x2 at the point x = 3 is given by dy/dx.

Differentiate the curve with respect to x.

y = (x + 4)/x2

dy/dx = (x + 4)(1/x2)' + (1/x2)(x + 4)'

dy/dx = -2(x + 4)/x3 + (1/x2)(1)

⇒ dy/dx = -2(x + 4)/x3 + (1/x2)

When, x = 3, dy/dx = -2(3 + 4)/33 + (1/32) = - 11/27.

Therefore, the gradient of the curve at x = 3 is - 11/27.

b).

The cirve is y = (x + 4)/x2  and the tangent line is parallel to the line y = - 2.

y = - 2 is the horizontal line, so, slope of the line is zero.

Since, parallel lines have same slopes, the slope of the tangent line is aslo zero.

-2(x + 4)/x3 + (1/x2) = 0

[-2(x + 4) + x]/x3 = 0

-2x - 8 + x = 0

⇒ x = - 8.

Substitute x = - 8, in the given curve.

y = (- 8 + 4)/(- 8)2

= - 4/64

= - 1/16.

Therefore, the coordinates of the point, where the tangent is parallel to the line y + 2 = 0 is (- 8, - 1/16).