Sketch the Graph...

[nikchic5]

Sketch the graph of a function that satisfies all of the given conditions on its domain: f'(3.5)=f'(-3.5)=0, f'(x) is less than 0, if the absolute value of x is less than 3.5. f'(x) is greater than 0 is 3.5 is less than x is less than 7. f'(x)=-1 if the absolute value of x is greater than 7. f''(x)is less than 0 if -7 is lessthan x is less than 0. Inflection point where x=0.

Select the correct answer.
The choices are 4 different graphs.

Thank you so much for any help!

 

[tkhunny]

Sketch the graph of a function that satisfies all of the given conditions on its domain: f'(3.5)=f'(-3.5)=0, f'(x) is less than 0, if the absolute value of x is less than 3.5. f'(x) is greater than 0 is 3.5 is less than x is less than 7. f'(x)=-1 if the absolute value of x is greater than 7. f''(x)is less than 0 if -7 is lessthan x is less than 0. Inflection point where x=0.

Just draw the vague impressions that are suggested by the hints.

"f'(3.5)=f'(-3.5)=0"

Local min or max at x = 3.5 and x = -3.5

"f'(x) is less than 0, if the absolute value of x is less than 3.5"

Decreasing between x = -3.5 and x = 3.5
This suggests f(-3.5) is a local maximum and f(3.5) is a local minimum

"f'(x)=-1 if the absolute value of x is greater than 7"

This is a rather odd hint. Nevertheless, f(x) must go off to +BIG as x increases without bound in the negative direction (left of x = -7) and f(x) must go off to -BIG as x increases without bound in the positive direction (right of x = 7).

You figure out the last couple of clues.

 

[Gene]

It is hard without the graphs 'cause we can't tell how many are close.
f'(3.5)=f'(-3.5)=0, f'(x) is less than 0, if the absolute value of x is less than 3.5.
There is a maximum at x=-3.5 going down to a min at x=3.5
f'(x) is greater than 0 is 3.5 is less than x is less than 7. It goes up from 3.5 to 7
f'(x)=-1 if the absolute value of x is greater than 7.
It goes down in a straight 45° line from 7 and up. It goes up in a straight 45° line from -7 and down.
f''(x)is less than 0 if -7 is less than x is less than 0. Inflection point where x=0.
The slope changes at x=0

I'm too late but I'll (I hope) repeat what TSH said.

 

[soroban]

Hello, nikchic5!

Here's my impression of the graph . . .

Sketch the graph of a function that satisfies all of the given conditions on its domain:
(1) $\displaystyle f'(3.5)\,=\,f'(-3.5)\,=\,0.$
(2) $\displaystyle f'(x)\,<\,0\;$ if $\displaystyle |x|\,<\,3.5$
(3) $\displaystyle f'(x)\,>\,0\;$ if $\displaystyle 3.5\,<\,x\,<\,7$
(4) $\displaystyle f'(x)\,=\,-1\;$ if $\displaystyle |x|\,>\,7\;\;$ . . . Is this one correct?
(5) $\displaystyle f''(x)\,<\,0\;$ if$\displaystyle \,-7\,<\,x\,<\,0$
(6) Inflection point at $\displaystyle x\,=\,0$

(1) tell us that there are horizontal tangents at $\displaystyle x\,=\,3.5,\:-3.5$

          --*--   |   --*--
                  |
      - - - + - - + - - + - - -
          -3.5    |    3.5

(2) tell us that the graph goes downhill on the interval (-3.5, 3.5)

          --*-- 
                  \
                      --*--
      - - - + - - + - - + - - -
          -3.5    0    3.5

I suspect a relative maximum and a relative minimum.

(3) tells us that the graph is rising on the interval (3.5, 7).
(5) tell us that the graph is concave down on the interval (-7, 0).

          --*-- 
       /          \          /    
                      --*--
      - - - + - - + - - + - - -
          -3.5    0    3.5

I was right . . .
There is a relative maximum at $\displaystyle x\,=\,-3.5$; a relative minimum at $\displaystyle x\,=\,3.5$.
And there is an inflection point at $\displaystyle x\,=\,0$.


Now (4) tells us the graph is a straight line for $\displaystyle x\,<\,-7$ and $\displaystyle x\,>\,7$
This is possible with a contrived piece-wise function . . . not very satisfying.

If (4) had been: $\displaystyle f'(x)\,<\,0\,$ if $\displaystyle |x|\,>\,7$, the problem is more interesting.
The graph is going downhill for $\displaystyle x\,<\,-7$ and $\displaystyle x\,>\,7$

                --*-- 
       \     /          \          /    \
                            --*--
      - - + - - - + - - + - - + - - - + - -
         -7     -3.5         3.5      7

How is this possible?
There is a minimum at $\displaystyle x\,=\,3.5$
The curve rises until it reaches $\displaystyle x\,=\,7$ . . . then it goes downhill?
$\displaystyle \;\;$Wouldn't that require another relative maximum?
No, not if there is a vertical asymptote at $\displaystyle x\,=\,\pm7$
Even better, how about a horizontal asymptote: $\displaystyle y\,=\,0$ ?

The graph could look something like this:

          :             |            *:*
          :             |             :
          :      ***    |           * : *
          :             *         *   :     *
          :    *        |    ***      :         *
  - - - - + - - - + - - + - - + - - - + - - - - - -
   *     -7  *  -3.5    |    3.5      7
       *  :             |             :
        * : *           |             :
          :             |             :
         *:*            |

 

[nikchic5]

help

Is there anyother way it could look. Because out of my choices....none of them look like that. Sorry I really dont see it! Thanks so much for your time

 

[stapel]

Is there anyother way it could look. Because out of my choices....none of them look like that.

There are likely infinitely-many "correct" answers. You need to follow the reasoning to find the one (of the given four) that is one of those correct answers.

You've been given quite lengthy explanations of the general methodology, thinking, and processes. Where are you stuck?

Please reply with specifics regarding your work and reasoning. Thank you.

Eliz.

 

[nikchic5]

Question...

does the graph go from up to down or up to up or down to down? SOrry i dont understand. Thanks

 

[stapel]

does the graph go from up to down or up to up or down to down?

On which part? As the tutors have explained and illustrated, there are different "sections" of the graph to consider.

Eliz.

 

[Gene]

Re: o

OK, poor choice of words.
It goes down in a straight 45° line from 7 and up. It goes up in a straight 45° line from -7 and down.
Should read:
It slants down and right in a straight line from where x=+7. It slants up and left in a straight line from where x=-7. Both are parallel to y=-x.

You should be able to find something in each graph picture (but one) which doesn't match something we have said. When you do, forget that one. If the line after +7 isn't straight, that's not the graph.

 

[nikchic5]

one more

k one last question. Do you think the zero is at 4 or 5. (0,4) or (0.5) because there are two very simlar graphs? Thanks

 

[stapel]

Do you think the zero is at 4 or 5.

There is no specific information regarding the location of a zero. As was mentioned previously, there are many "right" answers. The point of this question is that you exercise your understanding of slopes, derivatives, inflection points, and the like, not that we pick the best choice of the four pictures you're looking at.

Eliz.

 

[nikchic5]

ok

ok well the correct answer is one of the choices. But the difference is where the zero is. Also the other major difference is that on has a bigger hump on both sides than the other. Thanks

 

[nikchic5]

and...

and sorry one has sharper edges at x=7 and x=-7

 

[stapel]

the correct answer is one of the choices.

No; "a" correct answer is one of the choices. There are infinitely-many graphs which would fit the listed conditions. The point is that you work through the function's listed properties, comparing them against the various pictures, and you determine which one is the best fit.

Eliz.

 

[Gene]

Do they both have an inflection at 0?

 

[nikchic5]

ok

one is the exact curve and the other one is the opposite. thanks