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

Summary### For the function g(x)=(1-2x)^2(x-3)? in

Full Text / Question### For the function g(x)=(1-2x)^2(x-3)? in **Science **

*For the function g(x)=(1-2x)^2(x-3)?*: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 …

__For the function g(x)=(1-2x)^2(x-3)?__
qwine2000 says

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.