All of them
A cubic function, continuous, differentiable.
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7
F(x)=[x^2]+1
f(x)=(x/|x-3|)+1; domain is all real numbers except 3. f(x)=(x/(|x-3|+1)); domain is all real numbers.
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
All of them
3.
At x = 3, the value of F(x) = 3x + 2 is the value 11, which graphs to the point (3, 11).
find f'(x) and f '(c)f(x) = (x^3-3x)(2x^2+3x+5
A cubic function, continuous, differentiable.
No, f(x) is not the inverse of f(x).
f(x) = x2 + 3x - 2 then f'(x) = 2x + 3 and then then f'(2) = 2*2 + 3 = 4+3 = 7
-2
F(x)=[x^2]+1
f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28