find the inverse prime for f(x)= 3+x^2+tan(pi/2*x) for a=3

Question

I am not even really sure where to start this problem. My teacher shows us the most generic examples and then assigns us very difficult problems. Please help me!!

Answer 1

f(x)=3+x²+tan(½πx).

If g(x)=f^-1(x), then f(g(x))=g(f(x))=x.

Assuming that we need g'(f(x)) (inverse prime) then differentiate g(f(x)) wrt x:

(d/dx)(g(f(x)))=g'(f(x))f'(x).

But g(f(x))=x, so g'(f(x))f'(x)=1 and:

g'(f(x))=1/f'(x).

f'(x)=2x+½πsec²(½πx), so g'(f(x))=1/(2x+½πsec²(½πx)).

g'(f(a))=1/(2a+½πsec²(½πa)). When a=3:

g'(f(3))=1/(6+½πsec²(3π/2)). cos(3π/2)=0 so sec(3π/2)→∞.

Therefore inverse prime=0.