If f(x) = x2 + 7, what is the equation for f'(x)?
1 answer:
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Answer:
{-1, 1/2, 2/3, 1}
Step-by-step explanation:
I find a graphing calculator to be a most useful tool for problems like this. (See attached.)
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First of all, observe that the constant is -2, so 0 is not a zero.
It is often useful to look at the sum of coefficients. Here, that is 0, indicating 1 is a zero.
You can divide that out, or simply continue by looking at the sum of coefficients with odd-degree terms negated. Then you have 6 +7 -4 -7 -2 = 0, indicating -1 is also a zero.
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At this point, it can be useful to divide out the factors you know. The synthetic division tableaus attached show that work. First is division by (x+1), then by (x -1).
The result of doing that is the quadratic ...
6x^2 -7x +2 = 0
This can be factored ...
(6x^2 -3x) -(4x -2) = 0 . . . . . . rewrite -7x and group term pairs
3x(2x -1) -2(2x -1) = 0 . . . . . . . factor each pair
(3x -2)(2x -1) = 0 . . . . . . the factorization of the remaining quadratic
The zeros of this are the values of x that make the factors zero: 2/3 and 1/2.
The zeros of the given function f(x) are x ∈ {-1, 1/2, 2/3, 1}.
