Please help me sketch this 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
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Please help me sketch this 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
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 xaxis.
Then at x=1 it changes concavityit faces upward, but it is still going down.
Then at x=2 it becomes horizontal until it reaches very close to positive infinity.
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.
Hello, ticbol!
Ouch!Quote:
The pesky negative sign bit you. :)