Question

suppose the function f is continuous at all real numbers and g(x)=f^(-1)(x) for all x. suppose the function g is also continuous at all real numbers and f(-2)=1 and f(5)= - 3. the intermediate value theorem guarantees that
o g(c)=0 for at least one c between - 2 and 5.
o g(c)=2 for at least one c between - 3 and 1.
o f(c)=0 for at least one c between - 3 and 1.
o f(c)=2 for at least one c between - 2 and 5.
question 5 of 10

Explanation:

Step1: Recall the Intermediate - Value Theorem

The Intermediate - Value Theorem 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 $f(c)=k$.

Step2: Analyze the given information

We know that $f$ is continuous on all real numbers, $f(-2)=1$ and $f(5)= - 3$. The inverse function $g(x)=f^{-1}(x)$.
If we consider the function $f$ on the interval $[-2,5]$, for any value $k$ between $-3$ and $1$, there exists a $c\in[-2,5]$ such that $f(c)=k$.
Since $g(x)$ is the inverse of $f(x)$, if we consider the domain and range swap property of inverse functions.
Let's assume we want to find a value of $c$ for $g(x)$. We know that if $y = f(x)$ and $x = g(y)$.
We want to find when $g(c)$ takes on certain values. Since $f$ is continuous on $[-2,5]$, and we know the values of $f$ at the endpoints.
If we consider the fact that for the inverse function $g$, we know that $g$ is also continuous (because $f$ is one - to - one and continuous).
We know that if we want to find a value such that $g(c) = 0$, we are essentially looking for a value of $x$ such that $f(x)=0$. Since $f$ is continuous on $[-2,5]$ and $f(-2)=1$ and $f(5)=-3$, by the Intermediate - Value Theorem, there exists at least one $c$ between $-2$ and $5$ such that $f(c)=0$.

Answer:

$f(c)=0$ for at least one $c$ between $-2$ and $5$.