Question

consider the following.

$f(x)=x^{2}+5x + 1,0,7,f(c)=15$

is $f$ continuous on the closed interval $0,7$?

yes

no

find the values of $f$ at $x = 0$ and $x = 7$.

$f(0)=$

$f(7)=$

determine whether the intermediate value theorem can be applied to $f$ on the closed interval $0,7$.

yes, the intermediate value theorem can be applied.

no, because $f$ is not continuous on the interval $0,7$.

no, because $f$ does not satisfy $f(0)

no, because $f(0)
eq f(7)$.

find the value of $c$. (if a value does not exist, enter na.)

$c=$

Explanation:

Step1: Check continuity of polynomial

Polynomials are continuous everywhere.

Step2: Compute f(0)

$f(0)=0^2+5*0+1=1$

Step3: Compute f(7)

$f(7)=7^2+5*7+1=49+35+1=85$

Step4: Check IVT conditions

f continuous on [0,7], 15 between 1 and 85.

Step5: Solve f(c)=15

$c^2+5c+1=15$ → $c^2+5c-14=0$ → $c=2$ (since -7 not in [0,7])

Answer:

A. Yes
f(0) = 1
f(7) = 85
A. Yes, the Intermediate Value Theorem can be applied.
c = 2