Question

consider function g. g(x) = { (1/2)^x +3, x< 0. -x^2 +2, x>_ 0

Answer 1

$\square$ The function is continuous. [False]

Both pieces of the function are continuous, so the overall continuity of $g(x)$ depends on continuity at $x=0$.

We have

$\displaystyle \lim_{x\to0^-} g(x) = \lim_{x\to0} \left(\frac1{2^x} + 3\right) = 1 + 3 = 4$

and

$\displaystyle \lim_{x\to0^+} g(x) = \lim_{x\to0} (-x^2+2) = 2$

The one-sided limits do not match, so $g$ is not continuous at $x=0$.

$\square$ As $x$ approaches positive infinity, $g(x)$ approaches positive infinity. [False]

$g(x)$ is a large negative number when $x$ is very large, so $g(x)$ is approaching negative infinity.

$\boxed{\checkmark}$ The function is decreasing over its entire domain. [True]

This requires $g'(x) \le 0$ on the entire real line. Compute the derivative of $g$.

$g'(x) = \begin{cases}-\ln(2)\left(\dfrac12\right)^x & x0 \end{cases}$

• $\left(\frac12\right)^x > 0$ for all real $x$, so $g'(x)$ whenever $x$.

• $x^2\ge0$ for all real $x$, so $-x^2\le0$ and $-x^2+2\le2$. Equality occurs only for $x=0$, which does not belong to $x>0$.

Whether the derivative at $x=0$ exists or not is actually irrelevant. The point is that $g(b) < g(a)$ if $b>a$ for all real $a,b$.

$\boxed{\checkmark}$ The domain is all real numbers. [True]

There are no infinite/nonremovable discontinuities, so all good here.

$\boxed{\checkmark}$ The $y$-intercept is 2. [True]

When $x=0$,

$g(0) = -0^2 + 2 = 2$