Let's simplify: f(x) = 3x-3x, so f(x) =0, so y=0
The graph is a line along the x-axis.
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The domain is what you choose it to be. You could, for example, choose the domain to be [3, 6.5] If the domain is the real numbers, the range is [-12.25, ∞).
the linearization of f(x) = x^4 + 3x^2 with a = -1 is L(x) = 4 - 10(x+1) the linearization formula is L(x) = f(a) + f'(a)*(x-a) when, a = -1 f(a) = (-1)^4 + 3(-1)^2 = 4 f'(x) = 4*x^3 + 6*x f'(a) = 4(-1)^3 + 6(-1) = -10 so, L(x) = (4) + (-10)*(x-(-1)) = 4 - 10(x+1)
f(x) = 1/3x3 f'(x) = ∫ 1/3x3 dx f'(x) = 1/3∫ x3 dx f'(x) = 1/3 * 3 * x2 f'(x) = x2
The domain could be the real numbers, in which case, the range would be the non-negative real numbers.
f x 3x plus 1 and g x x 2 -6 find fog 4 can not be solved.