F(x)=x²-8
g(x)=3x-2/2x-
f'(x)>g'(0)

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вопрос

First, let's find the derivatives of the functions f(x) and g(x):

f'(x) = 2x -
g(x) = (3x - 2) / (2x - 3)

To find g'(x), we can use the quotient rule:

g'(x) = [(2x - 3)(3) - (3x - 2)(2)] / (2x - 3)^
g'(x) = (6x - 9 - 6x + 4) / (2x - 3)^
g'(x) = -5 / (2x - 3)^2

Now let's find g'(0):

g'(0) = -5 / (2(0) - 3)^
g'(0) = -5 / (-3)^
g'(0) = -5 / 9

Now we need to find when f'(x) is greater than g'(0):

f'(x) > g'(0
2x - 8 > -5/9

Solving for x:

2x > 8 - 5/
2x > 72/9 - 5/
2x > 67/
x > 67/18

Therefore, f'(x) is greater than g'(0) when x is greater than 67/18.