13. does the ivt apply to the function $f(x)=-(\frac{1}{2})^{3 - x}-3$ on the interval $2,5$ for $f(c)=-4$?

Explanation:

Step1: Recall IVT conditions

The Intermediate - Value Theorem (IVT) states that if a function $y = f(x)$ is continuous on a closed interval $[a,b]$, and $k$ is a number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a,b)$ such that $f(c)=k$. First, we need to check the continuity of the function $f(x)=-\left(\frac{1}{2}\right)^{3 - x}-3$.

Step2: Analyze the continuity of the function

The exponential function $y = a^{u}$, where $a>0,a\neq1$ and $u$ is a real - valued function of $x$, is continuous for all real $x$. The function $u = 3 - x$ is a linear function and is continuous for all real $x$. The function $y=\left(\frac{1}{2}\right)^{3 - x}$ is continuous for all real $x$ since $a=\frac{1}{2}\in(0,1)$ and $u = 3 - x$ is continuous. Then $y=-\left(\frac{1}{2}\right)^{3 - x}-3$ is also continuous for all real $x$ as it is a linear combination of continuous functions. So, $f(x)$ is continuous on the interval $[2,5]$.

Step3: Calculate $f(2)$ and $f(5)$

Calculate $f(2)$: [ \begin{align*} f(2)&=-\left(\frac{1}{2}\right)^{3 - 2}-3\ &=-\frac{1}{2}-3\ &=-\frac{1 + 6}{2}\ &=-\frac{7}{2}=-3.5 \end{align*} ] Calculate $f(5)$: [ \begin{align*} f(5)&=-\left(\frac{1}{2}\right)^{3 - 5}-3\ &=-\left(\frac{1}{2}\right)^{-2}-3\ &=-4 - 3\ &=-7 \end{align*} ]

Step4: Check if $-4$ is between $f(2)$ and $f(5)$

We have $f(2)=-3.5$ and $f(5)=-7$. Since $-7<-4<-3.5$, the number $k = - 4$ is between $f(5)$ and $f(2)$.

Answer:

Yes