derivatives of graphs

J

JoeJ

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Jun 20, 2005
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I have this graph, which I'll try to sketch here that I'm trying to find derivatives for. Here's the graph...

&&&4&&&&&&&&&&&&&&----------
&&&&&&f&&&&&&&&&&/
-----3------&&&&&&&/ /&\
&&&&&&&&\&&&&&/&/&&&\
&\&&2&&&&\&&&/&/ &&&&\
&&\&&&&&&\&/&&/&&&&&&\
&&&&1&&g&/\&&/&&&&&&&&\
&&&&&\&&/&&\&/&&&&&&&&&\
&&&&0&&1&&&2&&&&&3&&4&&5

Hopefully someone understands...f starts from a straight line at y=3, ends at x=1, corner's down to (2,0), goes back up to (3,4), to form an upside down triangle, then forms a line at y=4 to the right. g starts at (-1,2), comes down to (1,0), goes back up to (3,3), comes back down to (5,0) to form another triangle. There is no better way I can explain this.

I'm trying to find the following...

1. If P(x) = f(x)g(x), find P'(0). I'm thinking f'(0)g'(0), which is 3, because y=1, and y=3 are the 2 points where f & g cross the y-axis, so 1*3 = 3

2. If Q(x) = (f(x)) / (g(x)), find Q'(4). I'm thinking here, (f(4)) / (g(4))) = 4/1.5...4 is y, where the line f meets y, and 1.5 is the y-value where x=4

3. If C(x) = f(g(x)), find C'(2). I'm thinking here, take g(2), the f(answer of g(2)). At y=2, there are 3 points....x=-1, x=1.5, x=2.5, x=3.5

Any help and explanation would be very much appreciated.
 
If P(x) = f(x)g(x), find P'(0). I'm thinking f'(0)g'(0)
Click to expand...

Product Rule: if P(x) = f(x)g(x), then P'(x) = f(x)g'(x) + f'(x)g(x).

f'(0)g'(0), which is 3, because y=1, and y=3 are the 2 points where f & g cross the y-axis, so 1*3 = 3
Click to expand...

f'(0) isn't f(0). f'(0) is the slope of f(x) where x = 0. Likewise for g.

If Q(x) = (f(x)) / (g(x)), find Q'(4). I'm thinking here, (f(4)) / (g(4)))
Click to expand...

That's Q(4), not Q'(4). Use the Quotient Rule, or the Product Rule, Chain Rule, and Power Rule if you don't like the Quotient Rule.

If C(x) = f(g(x)), find C'(2). I'm thinking here, take g(2), the f(answer of g(2)).
Click to expand...

That's C(2), not C'(2). Use the Chain Rule. By the way, instead of "answer of g(2)", you can just write "g(2)".
 
Ok, then do I check for rise/run then, and go from there? Like, for if P(x) - f(x)g(x), find P'(0), do I do rise/run for f(x), and g(x), then use the product rule?
 
Hello, JoeJ!

I think I know what's going on.
Let me give you <u>my</u> version of it . . .

f(x) starts from a straight line at y = 3, ends at x = 1,
corners down to (2,0), goes back up to (3,4), then forms a line at y = 4 to the right.

g(x) starts at (-1,2), comes down to (1,0), goes back up to (3,3), comes back down to (5,0).


I'm trying to find the following...

1. If P(x) = f(x)g(x), find P'(0).

2. If Q(x) = (f(x)) / (g(x)), find Q'(4).

3. If C(x) = f(g(x)), find C'(2).
Click to expand...
Code: 
       |          *(3,4)
       | (1,3)
   * * * * *     *
       |    *     
       |     *  *
       |      * 
 - - - + - - - * - -
             (2,0)
. . . . . . . . . . / . . . 3 . . . . .x <u><</u> 1
. . . f(x) . = . | -3x + 6. . 1 < x <u><</u> 3
. . . . . . . . . . \ . . . 4 . . . . .x > 3
Code: 
             |
             |      (3,3)
             |       *
      (-1,2) |      * *
         *   |     *   *
             *    *     *
   - - - - - + - * - - - * -
               (1,0)     (5.0)
. . . . . . . . . . ./ . . .- x + 1 . . . . . . -1 <u><</u> x <u><</u> 1
. . . g(x) . = . | . (3/2)x - (3/2) . . . .1 < x <u><</u> 3
. . . . . . . . . . .\ (-3/2)x + (15/2) . . 3 < x < 5


For (a), we need the Product Rule:
. . . P'(x) .= .f(x)·g'(x) + f '(x)·g(x)

We must find the values of those four expression on the right at x = 0
. . . and plug them in.

Edit: corrected that coordinate.
 
You are correct, soroban, except for the first graph, the point (5,0) should be (2,0)...how could 5 be in between 1 and 3? I think you know what is meant by it, you just typed in 5 by accident. Anyways, figuring out P'(x) = f(x)g(x), P'(0) = 3(-x + 1) = -3x + 3...do I have to solve all of them like that? I don't think that's the correct way, but that's what I came up with.

By the way, how did you get those to show up like that?
 
P(x) = f(x)g(x) for all x

==> P'(x) = f(x)g'(x) + f'(x)g(x) for all x

==> P'(0) = f(0)g'(0) + f'(0)g(0)

= 3( (2 - 0) / (-1 - 1) ) + 0(g(0))

= 3 * (-2/2)

= -3
 
Hello, JoeJ!

We have these two piece-wise functions.


. . . . . . . . . . / . . . 3 . . . . .x <u><</u> 1
. . . f(x) . = . | -3x + 6. . 1 < x <u><</u> 3
. . . . . . . . . . \ . . . 4 . . . . .x > 3


. . . . . . . . . . ./ . . .- x + 1 . . . . . . -1 <u><</u> x <u><</u> 1
. . . g(x) . = . | . (3/2)x - (3/2) . . . .1 < x <u><</u> 3
. . . . . . . . . . .\ (-3/2)x + (15/2) . . 3 < x < 5


For (a), we need the Product Rule: . P'(x) .= .f(x)·g'(x) + f '(x)·g(x)

Then: . P'(0) . = . f(0)·g'(0) + f '(0)·g(0)


Now we need the four values on the right side.

What is .f(0) .and .f '(0)?
. . . For x <u><</u> 1, .f(x) = 3 .and .f '(x) = 0
. . . Hence: . f(0) = 3, .f '(0) = 0

What is .g(0) .and .g'(0) ?
. . . For -1 <u><</u> x <u><</u> 1, .g(x) = -x + 1 .and .g'(x) -1
. . . Hence: . g(0) = 1, .g'(0) = -1

Plug them in: . P'(0) . = . (3)(-1) + (0)(1) . = . -3
 
Ok then, for 2, we use the quotient rule, like...

Q'(x) = (g(x) (d / dx) [ f (x) ] - f(x) (d / dx) [ g (x)] ) / ([g ( x ) ] ^2)

So, Q'(x) = (1.5 (d / dx) [4] - 4 (d / dx) [1.5] ) / (1.5 ^ 2)
Q'(4) = (g(4) * f'(4) - f(4) * g'(4)) / [g(4)] ^ 2) =(1.5 * ((4-0)/(3-2)) - 4 ((1.5 - 0)/(3 - 5))/1.5 ^ 2)) = (0 - 4(-1.5 / 2)) / 2.25 = (-3 / 2.25)

I'm not sure if that's right though.

Then, for 3...

f(g(2)) = f'(g'(2)) = f'(1.5) = 1.5

Don't think that's right, but, don't know what rule to use for it though.
 
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