[tex]\square[/tex] The function is continuous. [False]
Both pieces of the function are continuous, so the overall continuity of [tex]g(x)[/tex] depends on continuity at [tex]x=0[/tex].
We have
[tex]\displaystyle \lim_{x\to0^-} g(x) = \lim_{x\to0} \left(\frac1{2^x} + 3\right) = 1 + 3 = 4[/tex]
and
[tex]\displaystyle \lim_{x\to0^+} g(x) = \lim_{x\to0} (-x^2+2) = 2[/tex]
The one-sided limits do not match, so [tex]g[/tex] is not continuous at [tex]x=0[/tex].
[tex]\square[/tex] As [tex]x[/tex] approaches positive infinity, [tex]g(x)[/tex] approaches positive infinity. [False]
[tex]g(x)[/tex] is a large negative number when [tex]x[/tex] is very large, so [tex]g(x)[/tex] is approaching negative infinity.
[tex]\boxed{\checkmark}[/tex] The function is decreasing over its entire domain. [True]
This requires [tex]g'(x) \le 0[/tex] on the entire real line. Compute the derivative of [tex]g[/tex].
[tex]g'(x) = \begin{cases}-\ln(2)\left(\dfrac12\right)^x & x<0 \\\\ ? & x=0 \\\\ -2x & x>0 \end{cases}[/tex]
• [tex]\left(\frac12\right)^x > 0[/tex] for all real [tex]x[/tex], so [tex]g'(x)<0[/tex] whenever [tex]x<0[/tex].
• [tex]x^2\ge0[/tex] for all real [tex]x[/tex], so [tex]-x^2\le0[/tex] and [tex]-x^2+2\le2[/tex]. Equality occurs only for [tex]x=0[/tex], which does not belong to [tex]x>0[/tex].
Whether the derivative at [tex]x=0[/tex] exists or not is actually irrelevant. The point is that [tex]g(b) < g(a)[/tex] if [tex]b>a[/tex] for all real [tex]a,b[/tex].
[tex]\boxed{\checkmark}[/tex] The domain is all real numbers. [True]
There are no infinite/nonremovable discontinuities, so all good here.
[tex]\boxed{\checkmark}[/tex] The [tex]y[/tex]-intercept is 2. [True]
When [tex]x=0[/tex],
[tex]g(0) = -0^2 + 2 = 2[/tex]
