# Graph sketch

• Jun 13th 2009, 11:30 AM
AMaccy
Graph sketch
Given:

$\displaystyle f(0)=f(2)=0$
$\displaystyle f'(x) < 0$ if $\displaystyle x < 1$
$\displaystyle f'(1)=0$
$\displaystyle f'(x) > 0$ if $\displaystyle x > 1$
$\displaystyle f''(x) > 0$

How can I sketch a graph of the function $\displaystyle f$ given these characteristics? How would it look and how can I prove this?
• Jun 13th 2009, 12:38 PM
Amer
Quote:

Originally Posted by AMaccy
Given:

$\displaystyle f(0)=f(2)=0$
$\displaystyle f'(x) < 0$ if $\displaystyle x < 1$
$\displaystyle f'(1)=0$
$\displaystyle f'(x) > 0$ if $\displaystyle x > 1$
$\displaystyle f''(x) > 0$

How can I sketch a graph of the function $\displaystyle f$ given these characteristics? How would it look and how can I prove this?

the curve intersect the x-axis at two points (0,0) and (2,0) and you have f'(1) is a critical point because f'(1)=0
f is increasing in the interval x>1 $\displaystyle (1,\infty )$ and decreasing in the interval $\displaystyle (-\infty , 1)$ and the curve is concave up for all values of x

I think the curve is $\displaystyle x^2-2x$