# Sketch the graph

• Jan 28th 2007, 04:48 PM
qbkr21
Sketch the graph

f(x)=0 for x=0
f'(x)>0 for x<-2
f'(x)=0 for x>2
f'(x)=0 for x=-2
f'(x)<0 for -2<x<2
f"(x)<0 for x<1
f"(x)=0 for x=1
f" (x)>0 for 1<x<2

Thanks
• Jan 29th 2007, 02:13 AM
ticbol
Quote:

Originally Posted by qbkr21

f(x)=0 for x=0
f'(x)>0 for x<-2
f'(x)=0 for x>2
f'(x)=0 for x=-2
f'(x)<0 for -2<x<2
f"(x)<0 for x<1
f"(x)=0 for x=1
f" (x)>0 for 1<x<2

Thanks

I don't know how to draw in computers so I cannot show the sketch. But I can describe it.

f(x)=0,for x=0
That means the graph passes point (0,0).

f'(x)=0 for x=-2
That means the graph is horizontal at x = -2.

f"(x)=0 for x=1
That means the concavity of the graph changes at x = 1. There is an inflection point at x=1. The graph there needs not be horizontal, though.

f'(x)>0 for x<-2
The slope of the graph is positive to the left of x = -2.
So if at x = -2, the graph is horizontal, then there is maximum point at x = -2.

f'(x)<0 for -2<x<2
From x = -2 up to x = 2, the slope is negative.

f'(x)=0 for x>2
The graph is horizontal to the right of x = 2. Horizontal up to before infinity.

f"(x)<0 for x<1
The concavity of the graph is downwards to the left of x = 1.
That means from negative infinity up to x=1, the graph is like a vertical parabola that opens downward, whose vertex is at x = -2.

f" (x)>0 for 1<x<2
The concavity of the graph is upwards between x=1 and x=2. Yet, the slope there is still negative.

---------------------
Okay, if you are still with me, the graph is like a weak or tipsy number two.
From after negative infinity up to x=1 it looks like a parabola with its highest point at x = -2.
It passes through the origin (0,0), so the highest point of the "parabola" is above the x-axis.
Then at x=1 it changes concavity--it faces upward, but it is still going down.
Then at x=2 it becomes horizontal until it reaches very close to positive infinity.
• Jan 29th 2007, 05:17 AM
Soroban
Hello, qbkr21!

I totally agree with ticbol's analysis . . .

Quote:

Sketch the graph:

$\displaystyle f(x) = 0$

$\displaystyle \begin{array}{cccc}f'(x) > 0 & \text{for }x<-2 \\ f'(x) < 0 & \text{-}2 < x < 2 \\ f'(x) = 0 & \text{for }x = 2 \\ f'(x)=0 & \text{for }x>2\end{array}$

$\displaystyle \begin{array}{ccc}f"(x)<0 &\text{for }x<1 \\ f"(x)=0 & \text{for }x=1 \\ f"(x)>0 & 1<x<2 \end{array}$

Take the statements one at a time . . .

$\displaystyle f(0) = 0$: the graph contains the origin.

$\displaystyle f'(x) > 0\text{ for }x < -2$
. . To the left of $\displaystyle x = -2$, the graph is rising: $\displaystyle \nearrow$

$\displaystyle f'(x) < 0\text{ for }\text{-}2 < x < 2$
. . Between -2 and +2, the graph is falling: $\displaystyle \searrow$

$\displaystyle f'(2) = 0$
. . At $\displaystyle x = 2$, there is a horizontal tangent: $\displaystyle \rightarrow$

$\displaystyle f'(x) = 0\text{ for }x > 2$
. . To the right of $\displaystyle x = 2$, the graph is a horizontal line: $\displaystyle \rightarrow$

$\displaystyle f''(x) < 0\text{ for }x < 1$
. . To the left of $\displaystyle x = 1$, the graph is concave down: $\displaystyle \cap$

$\displaystyle f''(1) = 0$
. . At $\displaystyle x = 1$, there is an inflection point.

$\displaystyle f''(x) > 0\text{ for }1 < x < 2$
. . From $\displaystyle x = 1$ to $\displaystyle x = 2$, the graph is concave up: $\displaystyle \cup$

The graph looks something like this:
Code:

                      |                       |         *            |     *  :    *      |   *    :        *  |    1    2   - - - + - - - - - - * - - + - - + - - - - -       -2            |  *  :    :                       |    *:    :                       |    *    :                       |      *  :                       |          *  *  *  *                       |

Evidently, there is a horizontal tangent at $\displaystyle x = -2$,
. . but they didn't mention it.

• Jan 29th 2007, 09:32 AM
ticbol
Quote:

Originally Posted by Soroban
Hello, qbkr21!

I totally agree with ticbol's analysis . . .

Take the statements one at a time . . .

$\displaystyle f(0) = 0$: the graph contains the origin.

$\displaystyle f'(x) > 0\text{ for }x < -2$
. . To the left of $\displaystyle x = -2$, the graph is rising: $\displaystyle \nearrow$

$\displaystyle f'(x) < 0\text{ for }\text{-}2 < x < 2$
. . Between -2 and +2, the graph is falling: $\displaystyle \searrow$

$\displaystyle f'(2) = 0$
. . At $\displaystyle x = 2$, there is a horizontal tangent: $\displaystyle \rightarrow$

$\displaystyle f'(x) = 0\text{ for }x > 2$
. . To the right of $\displaystyle x = 2$, the graph is a horizontal line: $\displaystyle \rightarrow$

$\displaystyle f''(x) < 0\text{ for }x < 1$
. . To the left of $\displaystyle x = 1$, the graph is concave down: $\displaystyle \cap$

$\displaystyle f''(1) = 0$
. . At $\displaystyle x = 1$, there is an inflection point.

$\displaystyle f''(x) > 0\text{ for }1 < x < 2$
. . From $\displaystyle x = 1$ to $\displaystyle x = 2$, the graph is concave up: $\displaystyle \cup$

The graph looks something like this:
Code:

                      |                       |         *            |     *  :    *      |   *    :        *  |    1    2   - - - + - - - - - - * - - + - - + - - - - -       -2            |  *  :    :                       |    *:    :                       |    *    :                       |      *  :                       |          *  *  *  *                       |

Evidently, there is a horizontal tangent at $\displaystyle x = -2$,
. . but they didn't mention it.

Good morning, Soroban.

The 4th statement is actually, "f'(x) = 0 for x = -2". Not for x = 2. So the horizontal tangent at x = -2 is mentioned. That's why I said the vertex of the "parabola" is at x = -2.

The pesky negative sign bit you. :)
• Jan 29th 2007, 09:51 AM
Soroban
Hello, ticbol!

Quote:

The pesky negative sign bit you. :)
Ouch!