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Problem 1
Answer: year 4
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The break-even point is when the profit is $0. You neither earn money nor lose it.
Plug in p(x) = 0 and solve for x
p(x) = x^3 - 4x^2 + 5x - 20
x^3 - 4x^2 + 5x - 20 = p(x)
x^3 - 4x^2 + 5x - 20 = 0 ... replace p(x) with 0
(x^3 - 4x^2) + (5x - 20) = 0
x^2(x - 4) + 5(x - 4) = 0
(x^2+5)(x - 4) = 0
x^2+5 = 0 or x-4 = 0
The equation x^2+5 = 0 has no real solutions; however x-4 = 0 solves to x = 4.
So plugging x = 4 into p(x) will lead to p(x) = 0. Meaning that the company breaks even at year 4.
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Problem 2
Answer: choice B) between 2.5 and 3.0; between 4.0 and 4.5
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Look at the f(x) column. Specifically we are looking for the times when the numbers change from positive to negative, or vice versa. Somewhere in between this change, y will have to equal 0 at some point (at least once). Note how in row 2 and row 3, we have f(x) = 1.1 change to f(x) = -0.8; so the change is from positive to negative.
This means f(x) = 0 for some x value between x = 2.5 and x = 3.5. Also, the same kind of logic applies for the last two rows of the table as well pointing to another root between x = 4.0 and x = 4.5 (check out the attached images)
