Answers to Practice Final Exam Problems 2

1. Domain of  - certainly x must be postive for either the numerator or the denominator to exist. And since the denominator will never be zero for x>0, this is the only restriction - so the domain is x>0.

For the range, first look at the extremes of the domain -- as x tends to 0 from the right, the logarithm approaches -infinity and the denominator approaches zero - so the function approaches negative infinity. Also, as x goes to infinity, the denominator grows much faster than the numerator -- so the limit of the function as x goes to infinity is zero.

That means the range includes all negative numbers, but it also includes some positive numbers as well -- to get the exact range, we'll need to find the maximum of the function:

>	f:=ln(x)/sqrt(x); df:=diff(f,x);

>	solve(df=0,x);

>	subs(x=exp(2),f);

>	simplify(%);

This means the range of the function is all numbers less than or equal to 2/e.

2.  Let (x, )  be a random point on the graph...if the tangent line there goes through the origin, then the slope of the tangent line (value of the derivative) at that point is the same as the slope of the line from that point to (0,0).  In other words:

>	diff(exp(3*x),x)=(exp(3*x)-0)/(x-0);

>	solve(%,x);

So the point is (1/3,e).

3. A substitution integral: Let :

>	Int(exp(x)/(exp(x)+4),x=ln(1/3)..ln(3))=int(exp(x)/(exp(x)+4),x=ln(1/3)..ln(3));

4. The region:

>	plot({1/x^2,[[0,4],[4,4]],[[3,0],[3,4]]},x=0..4,0..5,color=blue,thickness=2);

The intersection of y=1/x^2 and y=4 is when x=1/2. So the area is:

>	area:=Int(4-1/x^2,x=1/2..3);

>	value(area);

5. Take the derivative:

>	f:=sqrt(x)*ln(x); df:=diff(f,x);

>	solve(df=0,x);

>	simplify(subs(x=%,f));

And since the derivative is negative for x less than 1/e^2 and positive for x greater than this, we have found the minimum, namely  -2/e.

6. This is a Riemann sum...to see for what, you have to factor an n out of the denominator:

  -- so it's the integral of 1/(1+x) from 0 to 1:

>	Limit(Sum(1/(n+k),k=1..n),n=infinity)=limit(sum(1/(n+k),k=1..n),n=infinity);

Note:

>	Int(1/(1+x),x=0..1)=int(1/(1+x),x=0..1);

7. To find the inverse function, solve the equation x=f(y) for y.

>	solve(x=exp(y)-exp(-y),y);

So there are two possible inverses, depending on whether we want the value of the function to be positive or negative.

8. The area is:

>	area:=int(exp(2*x),x=0..k);

>	solve(area=3,k);

So this is the value of k we seek.

9. Both ln(x) and  are defined only for x>0 -- we'll find the minimum of the difference as follows:

>	f:=sqrt(x)-ln(x);

>	diff(f,x);

>	solve(diff(f,x)=0);

And the derivative is positive when x>4 and negative when x<4...so this is the minimum. Therefore  for all x, since ln(4) is certainly less than 2 (so the minimum is positive).

10. Substitution integral -- let u=sin(x):

>	Int(Pi^sin(x)*cos(x),x)=int(Pi^sin(x)*cos(x),x);

11. First of all, the area between the parabola and the line y=9 is

>	bigarea:=Int(9-x^2,x=-3..3);

>	value(bigarea);

And the area between the parabola and the line y=k is:

>	smallarea:=Int(k-x^2,x=-sqrt(k)..sqrt(k));

>	value(smallarea);

We want the small area to be half of the big area:

>	solve(value(smallarea)=1/2*value(bigarea),k);

12. Substitution integral: Let

>	Int(1/(sqrt(x)*(1+sqrt(x))),x=1..4)=int(1/(sqrt(x)*(1+sqrt(x))),x=1..4);

13. This function is defined only for x>0 -- and as x goes to zero,  goes to 0. As x goes to infinity it goes to 1. So we're left looking for the max:

>	f:=x^(1/x); df:=diff(f,x);

>	solve(df=0,x);

So the max is e^(1/e). Here's the graph:

>	plot(x^(1/x),x=0..17,color=blue,thickness=2);

>	limit(x^(1/x),x=infinity);