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question: consider the function f(x)=ln(3x + 8) over the interval 1,7. …

Question

question: consider the function f(x)=ln(3x + 8) over the interval 1,7. does the extreme - value theorem guarantee an absolute maximum and minimum for f(x) on this interval? select the correct answer below: yes no

Explanation:

Step1: Check function continuity and interval

The function $f(x)=\ln(3x + 8)$ is a natural - logarithm function. The domain of $y = \ln(u)$ is $u>0$. For $u = 3x+8$, when $x\in[1,7]$, $u=3x + 8>0$. The natural - logarithm function and the linear function $3x + 8$ are both continuous on their domains. So $y=\ln(3x + 8)$ is continuous on the closed interval $[1,7]$.

Step2: Recall the extreme - value theorem

The extreme - value theorem states that if a function $y = f(x)$ is continuous on a closed interval $[a,b]$, then $f(x)$ has both an absolute maximum and an absolute minimum on $[a,b]$. Since $f(x)=\ln(3x + 8)$ is continuous on $[1,7]$, the extreme - value theorem guarantees the existence of an absolute maximum and minimum for $f(x)$ on $[1,7]$.

Answer:

Yes