Is (a-3) a factor of (a^3-6a^2+29)?
1 answer:
If (a - 3) is a factor then (a -3) will be 0 when a = 3. If you put a = 3 into the given cubic, you should get 0. Doing that, does not give 0.
f(3) = 8 - 6*9 + 29
f(3) = 8 - 54 + 29
f(3) = - 17
As further "proof", here is the graph. Notice that the graph cuts the x axis around -2 , 2.5 and 4.2 (These are rough guesses). The factors are
(a+2)(a - 2.5)(a - 4.2)
f(3) = 8 - 6*9 + 29
f(3) = 8 - 54 + 29
f(3) = - 17
As further "proof", here is the graph. Notice that the graph cuts the x axis around -2 , 2.5 and 4.2 (These are rough guesses). The factors are
(a+2)(a - 2.5)(a - 4.2)
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Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.