Question

find the domain of f(x) = sqrt(4-e^2x)

Answer 1

The domain is the set of all x such that f(x) exists. The square root function will fail to exist for real x when the radicand (stuff under the radical sign) is negative. So, we require\[4-e^{2x} \ge 0\]here. We solve,\[4 \ge e^{2x}\]and taking logs of both sides,\[\log 4 \ge 2x\]and so \[\frac{\log 4}{2} \ge x\]; that is,\[x \le \frac{\log 4}{2}\]

Answer 2

Taking the log on both sides does not change the direction of the inequality since the log function is a monotonic increasing; that is, for \[x_1>x_2\]then\[\log x_1 > \log x_2\]

Answer 3

i am definitely confused

Answer 4

What part is confusing?

Answer 5

domain of f(x)=sqrt(4-e^2x)

-> 4-e^2x>=0
-> e^2x<=4
-> apply log with base 'e' in both side
2x<=loge (4)
-> x<= ( loge (4) )/2;means x can be 0 to ( loge (4) )/2;
note:-> here you have only square root function which you hav to worry about;because inside the square root thing should be non-negative,hence we took >=0 for that
*-> we need not to worry about e^2x function ...here x can be anything between -infinity to +infinity

Answer 6

-> when you apply log with base e on both side of inequality it can't change the inequality