How do you use the Intermediate Value Theorem to show that the polynomial function #x^3 - 2x^2 + 3x = 5# has a zero in the interval [1, 2]?

Question

Elijah Abram · Accepted Answer

Assuming you meant #f(x)=x^3-2x^2+3x-5#
Evaluate #f(1)# and #f(2)# and note that the function is continuous with values on both sides of #0# Note: #x^3-2x^2+3x=5# is a polynomial equation; it is not a polynomial function. Reference to an equation having a zero in an interval is meaningless. #f(x)=x^3-2x^2+3x# does not have a zero in the interval #[1,2]# That only leaves #f(x)=x^3-2x+3x-5# as the intended function #f(1)=-3# #f(2)= +1# Since #f(x)# is continuous in the interval #[1,2]# it must have have values for everything in the range #[-3,+1]# including zero.

Emma Aldridge · Answer

Here's one way to do it.

Let #f(x) = x^3-2x^2+3x#. (Needed because the intermediate value theorem is a theorem about functions .) Observe that the equation #x^3 - 2x^2 + 3x = 5# has a root (a solution) exactly when #f(x)=5# So the question now is to show that for at least one number #c#, in #[1,2]#, we get #f(c)=5#. #f# is continuous on #[1,2]# (Because it is a polynomial and they are continuous everywhere.) #f(1) = 2# and #f(2) = 6# #5# is between #f(1)# and #f(2)#, so by the intermediate value theorem, there is at least one number #c#, in #[1,2]#, for which #f(c)=5#. That is, the original equation has a solution.

William Abdalla · Answer

undefined To use the Intermediate Value Theorem to show that the polynomial function $x^3 - 2x^2 + 3x - 5$ has a zero in the interval $[1, 2]$, you need to demonstrate that the function changes sign within that interval. First, evaluate the function at the endpoints of the interval: $f(1)$ and $f(2)$. If the values have different signs, then by the Intermediate Value Theorem, there exists at least one value $c$ in the interval $[1, 2]$ such that $f(c) = 0$.

$f(1) = (1)^3 - 2(1)^2 + 3(1) - 5 = -3$

$f(2) = (2)^3 - 2(2)^2 + 3(2) - 5 = 1$

Since $f(1) = -3$ and $f(2) = 1$, and they have different signs, the Intermediate Value Theorem guarantees that there exists at least one value $c$ in the interval $[1, 2]$ such that $f(c) = 0$, which means the polynomial has a zero in the interval $[1, 2]$.