Use the intermediate value theorem to show that the polynomial function has a zero in the given interval.
f(x)=9x^4-3x^2+5x-1;[0,1]
There are a number of ways to do this, lets go to the idiot's guide to Math...
f(0)=-1
f(1)=9-3+5-1=10
so how can one get to 10 from -1 by not crossing the y=0 axis?
Not sure that is why I asked :)
The IVT is dependent on the fact that f(x) is continuous. That is, f(x) cannot get from -1 to 10 without being 0 somewhere on the way.
If f is not continuous, then there might be a hole at f=0, so there would be no guarantee that f(c)=0 for some 0<c<1.
Reread the intermediate value theorem, it concludes that one can't get to 10 from -1 with a continuous function without passing the y=0 axis. Often, the mean value, and intermediate value theorem are written in math texts by lawyer want-to-be types, so complex, it loses its meaning.
To use the intermediate value theorem to show that a polynomial function has a zero in a given interval, we need to show that the function changes sign in that interval. Here's how we can do it with the given polynomial function f(x) = 9x^4 - 3x^2 + 5x - 1 on the interval [0, 1]:
1. Evaluate f(0) and f(1):
- f(0) = 9(0)^4 - 3(0)^2 + 5(0) - 1 = -1
- f(1) = 9(1)^4 - 3(1)^2 + 5(1) - 1 = 10
2. Check the signs of f(0) and f(1):
- f(0) is negative (less than zero).
- f(1) is positive (greater than zero).
Since f(0) is negative and f(1) is positive, we have shown that the function f(x) changes sign from negative to positive on the interval [0, 1], which means there must be at least one zero of the function in that interval.