select all the correct locations on the graph. consider the given piece - wise function. $f(x)=\begin{cases}…

select all the correct locations on the graph. consider the given piece - wise function. $f(x)=\begin{cases}-(3x + 7);&x < - 3\\2x^{2}-16;&-3leq xleq3\\-(2^{x}-10);&x > 3end{cases}$ select the section(s) of the graph where the function is decreasing.

select all the correct locations on the graph. consider the given piece - wise function. $f(x)=\begin{cases}-(3x + 7);&x < - 3\\2x^{2}-16;&-3leq xleq3\\-(2^{x}-10);&x > 3end{cases}$ select the section(s) of the graph where the function is decreasing.

Answer

Explanation:

Step1: Analyze $f(x)=-(3x + 7)$ for $x < - 3$

The slope of the linear - function $y=-(3x + 7)=-3x - 7$ is $m=-3<0$. So it is decreasing for $x < - 3$.

Step2: Analyze $f(x)=2x^{2}-16$ for $-3\leq x\leq3$

The derivative of $y = 2x^{2}-16$ is $y^\prime=4x$. Set $y^\prime<0$, we get $4x<0$ or $x < 0$. So it is decreasing on the interval $[-3,0]$.

Step3: Analyze $f(x)=-(2^{x}-10)$ for $x > 3$

The derivative of $y = 2^{x}$ is $y^\prime=2^{x}\ln(2)>0$ for all $x$. Then the derivative of $y=-(2^{x}-10)=-2^{x}+10$ is $y^\prime=-2^{x}\ln(2)<0$ for all $x$. So it is decreasing for $x > 3$.

Answer:

The function is decreasing on the intervals $x < - 3$, $-3\leq x\leq0$, and $x > 3$. On the graph, this corresponds to the left - most blue line segment ($x < - 3$), the part of the red parabola from $x=-3$ to $x = 0$, and the right - most green line segment ($x > 3$).