suvadip said:If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).
But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?
The mean value theorem: if f is continuous and differentiable on [a, b] then there exist c in [a, b] such that [tex]f'(c)= \frac{f(b)- f(a)}{b- a}[/tex]. If f'(x)= 0 for all x, then for any a and b, [itex]\frac{f(b)- f(a)}{b- a}= 0[/tex] from which f(a)= f(b).suvadip said:If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).
But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?
A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is essentially the slope of the tangent line to the function at that point.
To prove that a function f(x) is constant, you need to show that its derivative, f'(x), is equal to 0 for all x values. This means that the function's rate of change is 0, indicating that it is not changing and therefore constant.
Knowing how to prove that a function is constant using derivatives is important in understanding the behavior and characteristics of functions. It also allows for the identification of constant functions, which are often used as a starting point for more complex mathematical concepts.
The derivative of a constant function f(x) = c, where c is any constant, is equal to 0. In other words, the derivative of a constant function is always 0.
For example, let's take the function f(x) = 5. To prove that this function is constant, we can take its derivative, which is f'(x) = 0. This shows that the function has a constant rate of change of 0, and therefore it is a constant function.