# Help with a problem?

## the f be a function that is continuous on the closed interval [2,8]. Given that f(2)=2 and f(8)=12, which of the following statements is guaranteed by the intermediate value theorem f(x) is linear 2.f(x) is always increaseing 3.f(x)=7 has at least one solution in the open interval (2,8) 4.f(5)=7 5.there is at least one maximum on the open interval (2,8)

Jun 28, 2018

The third one.

#### Explanation:

The third one.

In fact as $f \left(x\right)$ is continuous in $\left[2 , 8\right]$, and $f \left(2\right) = 2$, $f \left(8\right) = 12$, the intermediate value theorem states that for every $\overline{y} \in \left(2 , 12\right)$ there must be a value $\overline{x} \in \left(2 , 8\right)$ such that:

$f \left(\overline{x}\right) = \overline{y}$

As $7 \in \left(2 , 12\right)$, then there must be $\overline{x} \in \left[2 , 8\right]$ such that:

$f \left(\overline{x}\right) = 7$

On the contrary if you let:

$f \left(x\right) = \frac{3 {x}^{2} - 25 x + 44}{3}$

$f \left(2\right) = 2$

$f \left(8\right) = 12$

the function is continuous for $x \in \left[2 , 8\right]$ but is not linear, it's not strictly increasing and $f \left(5\right) \ne 7$.

$f \left(x\right) = \frac{5 x - 4}{3}$
then the hypotheses are still satisfied but $f \left(x\right)$has no maximum in $\left(2 , 8\right)$