Question

For `f(x) = x^2-3x+8`, what are the intervals on which increasing, decreasing, concave up, concave down and the x-coordinates of all inflection points?

Answer 1

The interval of decreasing is `x in (-oo,3/2)`. The interval of increasing is `x in (3/2,+oo)`. The curve is concave up
on the interval `(-oo,+oo)` and no inflection points.

Explanation:

Our function is `f(x)=x^2-3x+8`

This function is defined and continuous for `x in RR`

The first derivative is

`f'(x)=2x-3`

The critical point is

`f'(x)=0`, `=>`, `2x-3=0`, `x=3/2`

We build a variation chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,3/2)` `color(white)(aaaa)` `(3/2,+oo)`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaa)` `↘` `color(white)(aaaaaaaaaaa)` `↗`

The interval of decreasing is `x in (-oo,3/2)`

The interval of increasing is `x in (3/2,+oo)`

The second derivative is

`f''(x)=2`

As `f''(x)>0`, we have a minimum, ie the curve is concave up

on the interval `(-oo,+oo)`

There are no inflection points.

graph{x^2-3x+8 [-10, 10, -5, 5]}