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.
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7
y = f(x) = 2x2 - 3x + 10dy = (4x - 3) dxNote:It's possible that the function in the question may have been f(x) = (2x)2 - 3x + 10 .If that's the case, thendy = (8x - 3) dx
fx=3x-5 f=(3x-5)/x f2=(6x-10)/x /=divide
-29.2 F
The question is not clear. A function is defined by an equation and that requires an equals sign. there is no equals sign in the question. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals".
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, ∞).
3x minus 4f=? x=3 f=4
if f(x) = 3x - 10, then whatever is put (substituted) for x in the "f(x)" bit is substituted for x in the "3x - 10" bit. Thus f(2a) = 3(2a) - 10 = 6a - 10.
f(x) = (3x - 4)/5 f(0) = [3(0) - 4]/5 = -4/5
3-e = 4-f f-e = 4-3 f-e = 1
-32.8 degrees F