label Calculus
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

Consider the function f (x) = 5x2 - 9x - 1. Does the Intermediate Value Theorem guarantee that f has a root (also called a zero) on the interval 1 ≤ x ≤ 2 ? Explain thoroughly.

Jun 10th, 2015

Okay, so the intermediate value theorem (IVT) says that a function must have a root if it is continuous on an interval [a,b], and there a set of values c,d on the interval such that f(c) < 0 and f(d) > 0. In simply english, that means the following: if a function is continuous meaning that there are no "holes" in the graph anywhere or asymptotes; in short, if you can draw the function without lifting you pencil then the function is continuous. So if there are two points, one that makes f(x) negative meaning that the graph is below the x axis, and another that makes f(x) positive so it is above the x axis, if you have to connect these points with a continuous line it must cross the x axis, so there has to be a root there.
Now let's apply this theorem to the problem.
So let's try x = 1:
f(1) = 5*(1)^2-9*(1)-1 = 5-9-1 = -5

Now lets try x = 2:
f(2) = 5*(2)^2 - 9*(2) -1 = 1
So because f(x) goes from negative to positive and f(x) is continuous for x in [1,2], the intermediate value theorem says that there must be a root between x = 1 and x = 2.

Jun 10th, 2015

...
Jun 10th, 2015
...
Jun 10th, 2015
Oct 18th, 2017
check_circle