Question

existence of multiple zeros in exercises 87 and 88, explain why the function has at least two zeros in the interval (1,5). 87. f(x) = (x - 3)² - 2 88. f(x) = 2 cos x

Answer:

To solve the problem of explaining why the function \( f(x)=(x - 3)^2-2 \) has at least two zeros in the interval \([1,5]\), we can use the Intermediate Value Theorem and analyze the function's behavior.

Step 1: Find the values of the function at the endpoints and critical points

First, let's find the critical point of the function \( f(x)=(x - 3)^2-2 \). The function is a quadratic function in the form \( y = a(x - h)^2+k \), where \( a = 1 \), \( h = 3 \), and \( k=-2 \). The vertex (critical point) of the parabola is at \( x = 3 \).

Now, we calculate the value of the function at \( x = 1 \), \( x = 3 \), and \( x = 5 \):

• For \( x = 1 \):

\( f(1)=(1 - 3)^2-2=(-2)^2-2 = 4 - 2=2 \)

• For \( x = 3 \):

\( f(3)=(3 - 3)^2-2=0 - 2=-2 \)

• For \( x = 5 \):

\( f(5)=(5 - 3)^2-2=(2)^2-2 = 4 - 2=2 \)

Step 2: Apply 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 any number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in the interval \((a,b)\) such that \( f(c)=k \).

1. Consider the interval \([1,3]\):

The function \( f(x)=(x - 3)^2-2 \) is a polynomial function, and all polynomial functions are continuous everywhere. So, \( f(x) \) is continuous on \([1,3]\). We have \( f(1) = 2 \) and \( f(3)=-2 \). Since \( 0 \) is between \( f(1) = 2 \) and \( f(3)=-2 \) (i.e., \( f(3)<0<f(1) \)), by the Intermediate Value Theorem, there exists at least one number \( c_1\in(1,3) \) such that \( f(c_1)=0 \).

1. Consider the interval \([3,5]\):

The function \( f(x) \) is also continuous on \([3,5]\) (because it is a polynomial). We have \( f(3)=-2 \) and \( f(5) = 2 \). Since \( 0 \) is between \( f(3)=-2 \) and \( f(5)=2 \) (i.e., \( f(3)<0<f(5) \)), by the Intermediate Value Theorem, there exists at least one number \( c_2\in(3,5) \) such that \( f(c_2)=0 \).

Since we have found two distinct intervals \((1,3)\) and \((3,5)\) within \([1,5]\) where the function has a zero, the function \( f(x)=(x - 3)^2-2 \) has at least two zeros in the interval \([1,5]\).

For the function \( f(x) = 2\cos x \) (assuming there is a typo and it should be \( f(x)=2\cos x \) with the interval \([1,5]\) in radians, since \( 1\) and \( 5\) are likely in radians as \( 5\) radians is about \( 286.48^\circ\)):

Step 1: Analyze the cosine function

The cosine function \( y=\cos x \) is continuous everywhere. So, \( f(x) = 2\cos x \) is also continuous everywhere.

We know that the cosine function has a period of \( 2\pi\approx6.28 \). The interval \([1,5]\) is within one period of the cosine function.

The cosine function \( y = \cos x \) has zeros at \( x=\frac{\pi}{2}\approx1.57 \) and \( x=\frac{3\pi}{2}\approx4.71 \), which are both within the interval \([1,5]\) (since \( 1<1.57<5 \) and \( 1<4.71<5 \)).

Step 2: Find the zeros of \( f(x) = 2\cos x \)

To find the zeros of \( f(x) = 2\cos x \), we set \( f(x)=0 \):
\( 2\cos x=0 \)
Divide both sides by \( 2 \):
\( \cos x = 0 \)
The solutions to \( \cos x=0 \) in the interval \([1,5]\) are \( x = \frac{\pi}{2}\) and \( x=\frac{3\pi}{2}\) (since \( \frac{\pi}{2}\approx1.57\in[1,5] \) and \( \frac{3\pi}{2}\approx4.71\in[1,5] \)). So, the function \( f(x) = 2\cos x \) has at least two zeros in the interval \([1,5]\).

Final Answer (for \( f(x)=(x - 3)^2-2 \)):

The function \( f(x)=(x - 3)^2-2 \) is continuous on \([1,5]\). \( f(1) = 2 \), \( f(3)=-2 \), and \( f(5) = 2 \). By the Intermediate Value Theorem, there is a zero in \((1,3)\) (since \( f(1)>0 \) and \( f(3)<0 \)) and a zero in \((3,5)\) (since \( f(3)<0 \) and \( f(5)>0 \)), so \( f(x) \) has at least two zeros in \([1,5]\).

Final Answer (for \( f(x) = 2\cos x \)):

The function \( f(x)=2\cos x \) is continuous on \([1,5]\). The equation \( 2\cos x = 0 \) simplifies to \( \cos x=0 \), whose solutions \( x=\frac{\pi}{2}\approx1.57 \) and \( x=\frac{3\pi}{2}\approx4.71 \) lie in \([1,5]\), so \( f(x) \) has at least two zeros in \([1,5]\).