Example 1: # 41 (page 173)
| > | restart; |
The function of interest in this problem is:
| > | y := a*sqrt(x) + b/sqrt(x); |
| > |
The condition that the function pass through the point (4, 13) means that y = 13 when x = 4:
| > | eq1 := eval( y=13, x=4 ); |
Read this command as: eval uate the equation y=13 when x=4 and assign the result to the name eq1 .
The square roots of 4 in this equation are a little surprising. To ask Maple to simplify this result to what we would expect, use
| > | eq1a := simplify( eq1 ); |
| > |
For the point ( 4, 13 ) to be an inflection point for this function the second derivative must change sign at x=4. The (first and) second derivative are:
| > | Dy := diff( y, x ); |
| > | D2y := diff( Dy, x ); |
These results are messier than they should be. Maple will simplify the expressions if we ask appropriately:
| > | Dya := simplify( Dy ); |
| > | D2ya := simplify( D2y ); |
| > |
From these equations it is easily seen that
is a singular point for the function,
will be a stationary point, and
is a possible inflection point. We could use this information together with
eq1a
, but let's see how we could use Maple to solve both equations.
| > | eq2 := eval( D2ya=0, x=4 ); |
| > | eq2a := simplify( eq2 ); |
| > |
The solution to equations eq1a and eq2a can be found with the command:
| > | sol := solve( { eq1a, eq2a }, { a, b } ); |
It is important that both the system of equations and the set of unknowns are specified as sets , i.e., with delimiters { } .
| > |
To find the function defined by these values of the parameter:
| > | Y := eval( y, sol ); |
| > |
But, we are not done! We know only that this function has a potential inflection point at (4,13). To verify that this point is an inflection point it is necessary to verify that the second derivative changes sign at
:
| > | D2Y := simplify( diff( Y, x,x ) ); |
For x>0, the denominator is positive. For 0 <
< 4,
< 0 so the second derivative is positive. For
> 4,
> 0 so the second derivative is negative.
As an alternative, plot the second derivative on an interval containing
:
| > | plot( D2Y, x=3..5 ); |
By either line of reasoning, this function changes from concave up to concave down at
and (4,13) is an inflection point.
| > | plot( Y, x=0..50, view=[0..50,0..50] ); |
| > |
Note: Much of this problem could also be done using the LinearMotion Maplet. It would, however, require a fair amount of copying and pasting (or writing by hand).
| > |
).
). (Note that
and
are not in the domain of the function, these points will need to be dealt with separately.) This can be completed using the LinearMotion Maplet or by direct computation within Maple.
= 0.643501:
), the endpoints are not critical points. However, the behavior of the function does need to be considered at these points. This will be accomplished by evaluating the one-sided limits:
and
. Use either the LimitCheck Maplet or explicit computations within Maple to evaluate these limits.
tells us that there is no global maximum value for this function on (0,
).
,
), (
,
), and (
,
).