The expression fxfxf means f(f(x)f(x)), where f(x) is a function of x. This is not equivalent to f cubed (f^3(x)), which would mean f(f(f(x))). In fxfxf, the function f(x) is applied twice to the input x, whereas in f cubed, the function is applied three times. The two expressions are different due to the number of times the function is applied to the input.
Let f(x)=abs(x) , absolute value of x defined on the interval [5,5] f(x)= |x| , -5 ≤ x ≤ 5 Then, f(x) is continuous on [-5,5], but not differentiable at x=0 (that is not differentiable on (-5,5)). Therefore, the Mean Value Theorem does not hold.
Even polynomial functions have f(x) = f(-x). For example, if f(x) = x^2, then f(-x) = (-x)^2 which is x^2. therefore it is even. Odd polynomial functions occur when f(x)= -f(x). For example, f(x) = x^3 + x f(-x) = (-x)^3 + (-x) f(-x) = -x^3 - x f(-x) = -(x^3 + x) Therefore, f(-x) = -f(x) It is odd
You would have been given a function for f(x) and another function for g(x). When you want to find f(g(x)), you put the function for g(x) wherever x occurs in f(x). Example: f(x)=3x+2 g(x)=x^2 f(g(x))=3(x^2)+2 I'm not sure what you mean by address domain and range. They depend on what functions you're given.
You must provide a complete equation for the function. Do you mean f(x) = 1.5x + 7.6 or something else?
I think you might mean f(x)+2? Or do you mean f(x+2)? Either way it depends on what f(x) is.
If you mean: f(x) = x4 - 3x3 + 5x2 / x2 then: f(x) = x4 - 3x3 + 5 ∴ f'(x) = 4x3 - 9x2 If you mean: f(x) = (x4 - 3x3 + 5x2) / x2 then: f(x) = x2 - 3x + 5 ∴ f'(x) = 2x - 3
For a function of only one variable it mean the derivative with respect to that variable. Thus, f'(x) = df(x)/dx. Occasionally, it can also refer to a variation of a function. For example, a family of functions, f(x), f'(x), f''(x) and so on.
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The expression fxfxf means f(f(x)f(x)), where f(x) is a function of x. This is not equivalent to f cubed (f^3(x)), which would mean f(f(f(x))). In fxfxf, the function f(x) is applied twice to the input x, whereas in f cubed, the function is applied three times. The two expressions are different due to the number of times the function is applied to the input.
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
Let f(x)=abs(x) , absolute value of x defined on the interval [5,5] f(x)= |x| , -5 ≤ x ≤ 5 Then, f(x) is continuous on [-5,5], but not differentiable at x=0 (that is not differentiable on (-5,5)). Therefore, the Mean Value Theorem does not hold.
While no set of rules can handle differentiating every expression, the following should help. For all of the following, assume c and n are constants, f(x) and g(x) are functions of x, and f'(x) and g'(x) mean the derivative of f and g respectively. Constant derivative rule:d/dx(c)=0 Constant multiple rule:d/dx(c*f(x))=c*f'(x) Sum and Difference Rule:d/dx(f(x)±g(x))=f'(x)±g'(x) Power rule:d/dx(xn)=n*xn-1 Product rule:d/dx(f(x)*g(x))=f'(x)*g(x) + g'(x)*f(x) Quotient rule:d/dx(f(x)/g(x))=(f'(x)*g(x)-g'(x)*f(x))/f(x)² Chain rule:d/dx(f(g(x))= f'(g(x))*g'(x)
3 x f x f x f x f x f x f = 3f6
f(x) is the same thing as y= example: f(x)=2x+3 OR y=2x+3
lil tj
Suppose you wish to differentiate x/f(x) where f(x) is a differentiable function of x, and writing f for f(x) and f'(x) for the derivative of f(x), d/dx (x/f) = [f - x*f']/(f2)