Question

let f be the function given by f(x)=x + tan(\frac{x}{5})-10. the intermediate value theorem applied to f on the closed interval 12, 15 guarantees a solution in 12, 15 to which of the following equations?

a f(x)=-10

b f(x)=0

c f(x)=4

Explanation:

Step1: Recall 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: Evaluate \(f(12)\) and \(f(15)\)

First, find \(f(12)=12+\tan(\frac{12}{5})- 10=2+\tan(2.4)\). Since \(\tan(2.4)\approx - 3.42\), then \(f(12)=2 - 3.42=-1.42\).
Next, find \(f(15)=15+\tan(3)-10 = 5+\tan(3)\). Since \(\tan(3)\approx0.14\), then \(f(15)=5 + 0.14 = 5.14\).

Step3: Check the values in options

We need to find a value \(k\) such that \(k\) is between \(f(12)\approx - 1.42\) and \(f(15)\approx5.14\).
For option A, \(f(x)=-10\), and \(-10\) is not between \(-1.42\) and \(5.14\).
For option B, \(f(x) = 0\), and \(0\) is between \(-1.42\) and \(5.14\).
For option C, \(f(x)=4\), but we need to first check if it's guaranteed. Since we know the Intermediate - Value Theorem guarantees a value between the function values at the endpoints of the interval, and we found the range of \(f(x)\) on \([12,15]\) from the above calculations. Here, while \(4\) is also between \(-1.42\) and \(5.14\), the most common use - case of the Intermediate - Value Theorem is to find a root (where \(f(x)=0\)).

Answer:

B. \(f(x)=0\)