Question

How do I do question 2 & 3 ? Please help thank you

Answer 1

2.

Given

• y=2x(2 -3x)

Find

• y·y'' +x·y' -16 in simplest form

Solution

It is convenient to expand the expression for y to ease determination of derivatives.

... y = 4x -6x²

... y' = 4 -12x

... y'' = -12

Then the differential expression can be written as

... (4x -6x²)(-12) +x(4 -12x) -16

... = -48x +72x² +4x -12x² -16

... = 60x² -44x -16

3.

Given

• y = 16/x +x³/3

Find

• the turning points
• the extreme(s)

Solution

The derivative is

... y' = -16x^-2 + x^2

This is zero at the turning points, so

... -16/x^2 +x^2 = 0

... x^4 = 16 . . . . . . . . . multiply by x^2, add 16

... x^2 = ±√16 = ±4

We're only interested in the real values of x, so

... x = ±√4 = ±2 . . . . . . . x-values at the turning points

Then the turning points are

... y = 16/-2 +(-2)³/3 = -8 +-8/3 = -32/3 . . . . for x = -2

... y = 16/2 + 2³/3 = 8 +8/3 = 32/3 . . . . . . . for x = 2

The maximum is (-2, -10 2/3); the minimum is (2, 10 2/3).