Intermediate Value Theorem Solution 3

Processing math: 100%

SOLUTION 3: We are given the equation

$x^3+2x-5=0$ . Let function

$f(x)=x^3+2x-5 \ \ \ \ and \ choose \ \ \ \ m=0$ This function is continuous for all values of

$x$ since it is a polynomial. To establish an appropriate interval consider the graph of this function. (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.)

tex2html_wrap_inline125

Note that

$f(1)= (1)^3+2(1)-5= -2<0 \ \ \ \ and \ \ \ \ f(2)= (2)^3+2(1)-5=5>0$
so that

$f(1)=-2 < m < 5=f(2)$
i.e.,

$m=0$ is between

$f(1)$ and

$f(2)$ . Now choose the interval to be

$\ [1, 2]$ .

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

$c$ in the interval

$[1, 2]$ which satisfies

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

$c^3+2c-5=0$ and the equation is solvable.

Click HERE to return to the list of problems.