Here, I'm given a function, and I need to know where f(x) = 0. The possible answers are -1, 0, 1, and 2. -1 should be the correct answer.
I thought I'd set the numerator to zero and get x that way, but that doesn't work because the powers are different.
Here is the counter again:
f(x) = x^3-3x^2+4x+8
Thanks in advance.
(1 votes)
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Your plan to set the counter equal to zero is correct. But as you said by the many different exponents you can not simply clip pq formula / midnight formula / x or apply similar “tricks”.
The official answer from school is: you need to guess the first zero. For this, in any order, you set the possible results (in the school actually always x=-2, x=-1, x=0, x=1 or x=2, here even indicated) and see if the result is 0.
For example: f(0)=0^3-3*0^2+4*0+8=8 is not 0 so 0 is not the correct solution, so one continues to try.
f(-1)=(-1)^3-3*(-1)^2+4*(-1)+8=-1-3-4+8=0 is a zero point.
If you want to know more zero points you expect:
F(x)/(x-zero), i.e. divided in writing
(x^3-3x^2+4x+8)/(x-(-1)=
(x^3-3x^2+4x+8)/(x+1)=x^2-4x+8
-(x^3+x^2)
-(-4x^2-4x)
-(8x+8)
By the way, this is called linear factor decomposition.
(x^2-4x+8)*(x+1)=x^3-3x^2+4x+8
From “x^2-4x+8” one can now theoretically determine the further zeros. Practically, there are complex numbers, so let’s do it.
If you now think of guessing first zero is doof must go differently, well it’s all right: there are approximation procedures like regular falsi or Newton procedures. There you start with guessed values, but the results are then used to determine the next values with which you would argue. That’s how you get closer. Super for computers, in the head or with a simple calculator, this is a senseless pastime. If the calculator has a ANS button, it’s better, but as already mentioned in the school, this is NOT required.
What if the denominator for the zero position -1 in the counter becomes zero? Let us take the denominator: x^2+3x+2. If linear factor decomposition is effected with the denominator, i.e. x^2+3x+2 =(x+1)*(x+2), then f(x) can be converted to ((x^2-4x+8)*(x-(-1)))/(x+2)*(x-(-1)). So (x-(-1)) would shorten away, so nothing contributes to the course of the function. Ergo would not exist. For this, one would have found a simpler way of writing to the zero point of f(x), because f(x) would then be simplified (x^2-4x+8)/(x+2) and would then be expected to continue.
A break is zero if the nenner is 0, at least if the counter is uneven. In doubt simply use and calculate all possibilities
A break is 0 if the counter is equal to zero and the denominator is equal to 0
Hello.
The denominator must not be 0, otherwise the division is invalid. So we can ignore it and we just have to focus on the counter.
Otherwise, inserting is the simplest method to find the zero point(s). Although you could do something combinatorial
x3 – 3×2 + 4x + 8 = 0
x3 – 3×2 + 4x = -8
This already shows that we cannot seek a positive number. Because if x was positive, then x3 + 4x > -3×2 would be. But it must be negative (-8). If -1 is the only negative number in the selection, this can only be the correct answer.
Often it is faster to use. 😉
LG
After this, it is necessary to check whether the denominator is also 0 at the counter zeroing point. Otherwise the break is not 0
Could one do, but then none of the answers would be correct, which one can exclude if one should cross the correct answer. 😉
But yes, if you want to check everything for accuracy, your objection would of course be right!