Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 15

You need to determine where the function increases or decreases, hence, you need to verify where f'(x)>0 or f'(x)<0.
You need to determine derivative of the function:
f'(x) = 2e^(2x) - e^(-x)
Putting f'(x) = 0, yields:
2e^(2x) - e^(-x) = 0
2e^(2x) - 1/(e^x) = 0 => 2e^(3x) - 1 = 0 => e^(3x) = 1/2 => e^x = root(3)(1/2)
x = ln root(3)(1/2)
Hence, the function decreases for x in (-oo, ln root(3)(1/2)) and the function increases for x in ( ln root(3)(1/2), oo) .
b)You need to remember that the function reaches it's extrema for x, such as f'(x) = 0.
The function reaches it's minimum point at x = ln root(3)(1/2).
c) The function is concave up for f''(x)>0 and concave down for f''(x)<0.
Evaluating f''(x) yields:
f''(x) = 4e^(2x) + e^(-x)
4e^(2x) + e^(-x) = 0=> 4e^(2x) + 1/(e^x) = 0 => 4e^(3x) + 1 = 0
e^x = root(3)(-1/4) = -root(3)(1/4) impossible since e^x>0 for all x in R.
Hence, the function has no inflection points and it is concave up for x in R , since f''(x) = 4e^(2x) + e^(-x) > 0 for all x in R .