help pleasE??

[mathhelp1a]

find all extema in the interval (-4, -2)
if y= 1 / x+1

how to differentiate?

i got -1/3 max
-1 min
am i right?

 

[chrisr]

Whether the graph is y = 1/(x+1) or (1/x)+1,
it is asymptotic, there are no extrema in the region x = -4 to x = -2.
y has upper and lower values in that region, but not extreme maximum and minimum values.

If you differentiate, you find the graph slope is always negative.
To differentiate, use the "quotient rule of differentiation",
or write the denominator as a power of -1 first instead.

You need to know the shape of this graph really.

 

[BigGlenntheHeavy]

You are right if the interval is [-4,-2], but wrong if the interval is (-4,-2).

f(-4) = -1/3 and f(-2) = -1, but neither are in the domain of f.

 

[chrisr]

nice one Glenn!

 

[mathhelp1a]

i meant [-4,-2]
-1/3 is in the domain

 

[chrisr]

yes, the line segment on the x axis from -4 to -2, inclusive.
In that case, x=-4 and x=-2 are in the domain and the maximum and minimum values
are those you calculated.
If the graph does not extend to the left of x=-4 or to the right of x=-2,
then you've got the maximum and minimum value of the curve segment
(are they called extrema too?).

 

[mathhelp1a]

what derivative test should i use, the first one ?

 

[chrisr]

In this case, calculating a derivative will not locate any maximums or minimums,
because differentiation locates local maximums and minimums, where a curve's slope changes sign.
The process discovers slopes = 0.
Your curve always has a negative slope everywhere.
If you bound it between x=-4 to x=-2, it is decreasing from f(-4) to f(-2),
so the maximum value is f(-4) and the minimum value is f(-2),
but these are not "local" maximums or minimums.
Do you see the difference?
In this case, we know the maximum and minimum because we know the slope is negative,
so the graph is continually decreasing from f(-4) to f(-2).

If you differentiate the function, you get f[sup:3dpewxtt]'[/sup:3dpewxtt](x) = -1/(1+x)[sup:3dpewxtt]2[/sup:3dpewxtt].
As the denominator is a square, it's +, so the slope is negative always, for all x.

 

[chrisr]

In terms of a "derivative test", you could say...

Since the derivative is negative always (having calculated it with the "quotient rule" or negative index),
we know the graph is heading downward from x=-4 to x=-2.
The derivative being negative means the graph is on the decrease in the direction of increasing x
from left to right along the x axis.
This is interpreted directly from knowing the shape of the graph.
Therefore f(-4) is the max and f(-2) is the min of the curve segment.