SOLUTION: Using the intermediate value theorem, show that the polynomial has a real zero between 1 and 2. f(x)=x^3-x-1

Question 689441: Using the intermediate value theorem, show that the polynomial has a real zero between 1 and 2.
f(x)=x^3-x-1
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Using the intermediate value theorem, show that the polynomial has a real zero between 1 and 2.
f(x)=x^3-x-1
------------------------
f(1) = 1-1-1 = -1
f(2) = 8-2-1 = 5
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Since f is a continuous function and is below the x-axis
when x = 1 but above the x-axis when x = 2, f must be zero
for some value between x = 1 and x = 2.
==========================================
cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
f(x) = x^3-x-1

f(1) = (1)^3-1-1

f(1) = 1-1-1

f(1) = -1

-------------------------------------------------------

f(x) = x^3-x-1

f(2) = (2)^3-2-1

f(2) = 8-2-1

f(2) = 5

Since f(1) = -1 and f(2) = 5, we have a change from a negative number to a positive number, so there's a zero between x = 1 and x = 2
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