i)state whether the function is even or odd

ii) find the x and y intercepts

iii) find the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points and concavity

i)state whether the function is even or odd

ii) find the x and y intercepts

iii) find the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points and concavity

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qwine2000

i) Neither, since, for example, g(1) = – 2 and g(- 1) = – 36.

ii) Let (1 – 2x)^2(x – 3) = 0

Then 1 – 2x = 0 or x – 3 = 0

The x-intercepts are at x = 1/2 or 3.

g(0) = 1^2*(-3) = – 3

The y-intercept is at y = – 3.

iii) g(x) = (2x – 1)^2(x – 3)

= (4x^2 – 4x + 1)(x – 3)

= 4x^3 – 16x^2 + 13x – 3

So dg/dx = 12x^2 – 32x + 13

and d^2g/dx^2 = 24x – 32

Setting dg/dx = 0, we get

12x^2 – 32x + 13 = 0

Since g(x) has 2 factors of (2x – 1), it follows that

dg/dx has one factor of (2x – 1).

The other factor must be (6x – 13).

So g(x) has stationary points at x = 1/2 or 13/6

The leading coefficient of g(x) is positive.

It follows that g(x) is increasing on (- infinity, 1/2) and (13/6, infinity).

Also, g(x) is decreasing on (1/2, 13/6).

Setting d^2g/dx^2 = 0, we get

24x – 32 = 0

24x = 32

x = 4/3

g(4/3) = (1 – 8/3)^2(4/3 – 3)

= (- 5/3)^2(- 5/3)

= – 125/27

g(x) has an inflection point at (4/3, – 125/27).

So g(x) is concave downwards for (- infinity, 4/3) and concave upwards for (4/3, infinity).

Note: There seems to be some confusion in the use of the terms ‘convex’ or ‘concave’. By ‘concave downwards’ I mean an upside-down bowl. By ‘concave upwards’ I mean a right-way-up bowl.