Intermediate Value Theorem Solution 1

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SOLUTION 1: We are given the equation

$3x^5-4x^2=3$ and the interval

$[0, 2]$ . Let function

$f(x)=3x^5-4x^2 \ \ \ \ and \ choose \ \ \ \ m=3$ This function is continuous for all values of

$x$ since it is a polynomial. (Please note that the graph of the function is not necessary for a valid proof, but the graph will help us understand how to use the Intermediate Value Theorem. On many subsequent problems, we will solve the problem without using the "luxury" of a graph.)

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Note that

$f(0)= 3(0)^5-4(0)^2= 2<3 \ \ \ \ and \ \ \ \ f(2)= 3(2)^5-4(2)^2=80>3$
so that

$f(0)=2 < m < 80=f(2)$
i.e.,

$m=3$ is between

$f(0)$ and

$f(2)$ .

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number

$c$ in the interval

$[0, 2]$ which satisfies

$f(c)=m$ i.e.,

$3c^5-4c^2= 3$ and the equation is solvable.

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