1. ## Function Question

a) Shade the area on the graph the region with points (x, y) which are solutions of the inequlity f(x) < y< g(x)

b) Use the graph to find the values of x for which the inequality in 3b is satisfied that is, the values of x for which f(x)<g(x). Do not solve the question algebraically. To receive any credit for it, your solution should be justified on the basis of your graph. Give your answer algebraically and in interval notation.

Can anyone help me with these two questions?

2. Hello, lemontea!

You have the curves labeled incorrectly in the graph.

a) Shade the area on the graph the region with points (x, y)
which are solutions of the inequality $f(x) \:< \:y \:< \:g(x)$

b) Use the graph to find the values of x for which $f(x)\,< \,g(x).$
Do not solve the question algebraically.
To receive any credit for it, your solution should be justified on the basis of your graph.

We want the region above $f(x)$ and below $g(x).$
Code:
              |  f(x) = 2x+1
*        |     /              * g(x) = (x-1)²
:       |    /
:*      |  /               *
::*     | /               *
::::*   |/              *
::::::::*           *
- -:-:-:-/+ - - * - - - - - - - -
::::::/ |
:::::/  |
::::/   |
:::/    |
::/     |
:/      |
/       |

(b) From the graph, it seems that $f(x) \,< \,g(x)$ when $x \:<\:0$

But it is known that a parabola "climbs" faster than a straight line.
Hence, further to the right, the parabola will intersect the line and exceed it.
From that point on, $f(x) \,<\,g(x).$

We can find that point algebraically.
Equate the two functions: . $(x-1)^2 \:=\:2x + 1\quad\Rightarrow\quad x^2-2x+1 \:=\:2x + 1$

. . $x^2-4x \:=\:0\quad\Rightarrow\quad x(x-4) \:=\:0\quad\Rightarrow\quad x\:=\:0,\:4$

Therefore, $f(x) \,<\,g(x)$ on the intervals $(-\infty,\,0)$ and $(4,\,\infty)$