f(x) = (e^(2x) - 1)/(ax) for x < 0, a ≠ 0

Question

If \(f(x) = \frac{e^{2x} - 1}{ax}\) for x < 0, a ≠ 0

f(x) = (e^2x - 1)/(ax) for x < 0, a ≠ 0

= 1 for x = 0

\(\frac{log(1 + 7x)}{bx}\) for x > 0, b ≠ 0

is continuous at x = 0, then find a and b.

Answer 1

f(x) = (e^2x - 1)/(ax)

Given, f is continuous at x = 0 and f(0) = 1