1. Identify the degree and describe the end behavior of the function: f(x)= 4^7-x^5+x^3−1.
2. Use the intermediate value theorem to show that the polynomial function has a real zero between the
numbers given: f(x)=x^4-4x^3-x+3; 0.5 and 1.
1. f(x)= 4^7-x^5+x^3−1
I suspect a typo, but as it stands, it is 5th degree, as x gets larger, f(large x) ----> + ∞
as x ----> large negatively, f(large negatives) ----> -∞
2. f(x)=x^4-4x^3-x+3
f(1/2) = 1/16 - 4(1/8) - 1/2 + 3 = 41/16
f(1) = 1 - 4 - 1 + 3 = -1
Since f(x) is continuous, and for f(1/2) its graph was above the x-axis
and at x = 1, its graph was below the x-axis, it must have crossed the
x-axis at least once, so there had to be at least one real zero between 1/2 and 1
it's an odd-degree polynomial, with positive leading coefficient. So, except for some possible bumps in the middle, it looks like f(x) = x^3
review the theorem.
f(0.5) > 0
f(1) < 0
so, for some 0.5 < c < 1, f(c) = 0