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(\frac12\right)^x & x<0 \\\\ ? & x=0 \\\\ -2x & x>0 \end{cases}`

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

•
`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`