Animated Solution for Mathematics - Functions: The largest interval lying in $(-\pi /2,\pi /2)$ for which the function, $f(x)=4^{-x^2}+{\cos }^{-1}(x/2-1)+\log (\cos x)$, is defined, is

To determine the domain of the function $f(x)=4^{-x^2}+{\cos }^{-1}(\frac{x}{2}-1)+\log (\cos x)$, we must identify the values of $x$ for which all three components are simultaneously defined. We treat this as an intersection problem:

The Relaxed Exponential

First, consider $f_1(x)=4^{-x^2}$. This is an exponential function where the exponent $-x^2$ is a polynomial.

Since exponential functions are defined for all real numbers, this component imposes no restrictions. Thus, the domain is:

The Strict Gatekeeper

Next, we examine $f_2(x)={\cos }^{-1}(\frac{x}{2}-1)$. The inverse cosine function ${\cos }^{-1}(u)$ is defined strictly for $u\in [-1,1]$.

We must enforce the following inequality:

Adding $1$ to all sides yields $0\le \frac{x}{2}\le 2$. Multiplying by $2$, we find the constraint:

The Logarithmic Boundary

Finally, we analyze $f_3(x)=\log (\cos x)$. The logarithm function requires its argument to be strictly positive.

We require $\cos x>0$. Within the context of the standard unit circle, this condition is satisfied in the interval:

The Final Convergence

To find the final domain, we calculate the intersection $D=D_1\cap D_2\cap D_3$. Since $D_1$ is the entire real line, we focus on the intersection of $[0,4]$ and $(-\frac{\pi }{2},\frac{\pi }{2})$.

Given that $\frac{\pi }{2}\approx 1.57$, which is less than $4$, the overlap is determined by the lower bound of $D_2$ and the upper bound of $D_3$.

The intersection starts at $0$ (inclusive) and ends at $\frac{\pi }{2}$ (exclusive, as $\cos (\frac{\pi }{2})=0$). Therefore, the domain is:

D =