Using differential to find the approximate value of 7.9 under the third root sign It's better to write it down on paper,

Using differential to find the approximate value of 7.9 under the third root sign It's better to write it down on paper,

When x tends to x0, f '(x0) is approximately [f (x) - f (x0)] / (x-x0)
F (x) = f (x0) + F '(x0) (x-x0)
This question: F (x) = 3 √ x, x = 7.9, x0 = 8, f '(x) = 1 / 3x ^ (- 2 / 3) substituting into the above formula, we get
f(7.9)=2-0.1*(1/3)*(1/4)

If y = ax and y = - B / X are decreasing functions on (0, + ∞), then y = ax & # 178; + BX is decreasing function on (0, + ∞)（

The functions y = ax and y = - B / X are decreasing functions on (0, + ∞)
A 0 is B

If a zero point of the function y = x ^ 3 + ax ^ 2 + BX + C is 1, then a + B + C=

A zero point of the function y = x ^ 3 + ax ^ 2 + BX + C is 1
That is to say, when x = 1, y = 0
So 1 + A + B + C = 0A + B + C = - 1
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