009A Sample Midterm 2, Problem 2

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The function is a polynomial and therefore continuous everywhere.

(a) State the Intermediate Value Theorem.

(b) Use the Intermediate Value Theorem to show that has a zero in the interval


Foundations:  
What is a zero of the function
        A zero is a value such that


Solution:

(a)

Step 1:  
Intermediate Value Theorem
        If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   is continuous on a closed interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]}
        and is any number between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(a)}   and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(b)} ,

        then there is at least one number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} in the closed interval such that

(b)

Step 1:  
First, is continuous on the interval since is continuous everywhere.
Also,

       

and

        .

Step 2:  
Since is between and
the Intermediate Value Theorem tells us that there is at least one number
such that
This means that has a zero in the interval


Final Answer:  
    (a)     See solution above.
    (b)     See solution above.

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