# Curve Sketching

• May 25th 2008, 09:48 AM
Macleef
Curve Sketching
Did I do the question properly?

Sketch a graph of a function f(x) that is differentiable and that satisfies the following conditions:
• $\displaystyle f'(x) > 0$, when $\displaystyle x < -3$ and $\displaystyle x > 1$
• $\displaystyle f'(x) < 0,$ when $\displaystyle -3 < x < 1$
• $\displaystyle f'(-3) = 0$ and $\displaystyle f'(1) = 0$
• $\displaystyle f'(-3) = 5$ and $\displaystyle f(1) = 1$
Legend:
Purple is f(x)
Green is f'(x)
• May 25th 2008, 10:16 AM
Moo
Hello,

No, it's false... the derivative is not positive in your sketch...
Actually, it's useless to draw f'(x). All that you have to remember is that if f'(x)>0, then f(x) will be increasing.

If f'(x)=0, there is a horizontal tangent line (here, it's g and g1 - see attachment), at A and B, which are turning points.

I've drawn a possible function, that satisfies the conditions :)

I hope it's clear enough... (sorry for the black line... couldn't do better)
• May 25th 2008, 12:44 PM
Soroban
Hello, Macleef!

I believe you have a typo in the problem . . .

Quote:

Sketch a graph of a function $\displaystyle f(x)$ that is differentiable
and that satisfies the following conditions:

$\displaystyle {\color{blue}[1]}\;\;f'(x) > 0\text{, when }x < \text{-}3\text{ and }x > 1$
$\displaystyle {\color{blue}[2]}\;\;f'(x) < 0\text{, when } \text{-}3 < x < 1$
$\displaystyle {\color{blue}[3]}\;\;f'(\text{-}3) = 0\text{ and }f'(1) = 0$
$\displaystyle {\color{blue}[4]}\;\;{\color{red}f(\text{-}3) = 5} \text{ and } f(1) = 1$

[1] says the graph is rising when $\displaystyle x < -3\text{ and } x > 1$
[2] says the graph is falling when $\displaystyle -3 < x < 1$
Code:

          :          :         * :  *        :      *       *  :    *    :    *     *    :        *  :  *   *      :          : *   - - - - + - - - + - + - - - - -         -3      0  1

[3] says the slope is 0 at $\displaystyle x = -3\text{ and }x = 1.$
Those are extreme points (or turning points or stationary points).
Code:

        --o--        :           :          :         * :  *        :    *       *  :    *    :  *     *    :        *  : *           :        --o--           :          :   - - - - + - - - + - + - - - - -         -3      0  1

[4] tell us exactly where those extreme points are.
We can now sketch the curve.

Code:

      (-3,5)    |           o      |          *       *    *    |     *        *  |        *     *            |        *                   *      *   *              |  o                   |    (1,1)   - - - - + - - - + - + - - - - -         -3      |  1                   |