Transcendental Functions
(Thomas-Finney, Section 6.1, page 456)
Problem 24
Let for x not 0. Find the inverse function, identifying its domain and range. As a check, writing the inverse function as g(x), show that both f(g(x))=x and g(f(x)) = x.
--------------------
> | f:=1/x^3; |
> | solve(y=f,x); |
We'll take the first one, because it's real!
> | finv:=subs(y=x,%[1]); |
> | simplify(subs(x=finv,f)); |
> | simplify(subs(x=f,finv),symbolic); |
So we're ok -- let's plot to make sure:
> | plot({f,x,finv},x=0..5,-1..5,color=blue,thickness=2,scaling=constrained); |
You can see that the two curves are reflections of one another (but it's not easy!).
Section 6.2, page 465, Problem 35
Find the derivative with respect to x of .
--------------------
Fun with the fundamental theorem:
> | assume(x>0); |
> | diff(int(ln(sqrt(t)),t=x^2/2..x^2),x); |
> | simplify(%); |
The little tildes "~" indicate that we have made an assumption about x.
Problem 57
Evaluate
.
> | restart; |
> | Int(2*ln(x)/x,x=1..2)=int(2*ln(x)/x,x=1..2); |
Section 6.3, page 472 , Problem 8
Solve
for y.
--------------------
> | solve(ln(1-2*y)=t,y); |
Problem 37
If
, compute the derivative of y with respect to x.
--------------------
> | eqn:=ln(y(x))=exp(y(x))*sin(x); |
> | solve(diff(eqn,x),diff(y(x),x)); |
Problem 48
Evaluate
.
--------------------
> | Int(exp(x/4),x=0..ln(16))=int(exp(x/4),x=0..ln(16)); |
Problem 77
a) Derive the linear approximation 1+x for at x = 0.
b) If x is in the interval [0, 0.2], estimate to 5 decimal places the error in using this approximation.
c) Graph and 1+x together for . Is one curve always above the other?
--------------------
(a) We do the usual tangent line idiom:
> | f:=exp(x); tanline:=simplify(subs(x=0,f)+subs(x=0,diff(f,x))*(x-0)); |
Let's plot first, so we can see where the error is worst:
> | plot({f,tanline},x=-0.2..0.5,color=blue,thickness=2); |
The line is always below the curve, and it's worst at x=0.2. So our upper bound for the error is
> | exp(0.2)-subs(x=0.2,tanline); |