So from a previous post I asked,
"f(x)=(1+x) what is f'(x) when x=0?"
and someone replied,
"the derivative of f(x)=x+1 is f'(x)=1 so your answer would just be 1"
Now my question is why!? it doesn't matter what x is? f'x when x=238423904 will also be 1? For some reason I thought that you would have to plug in 0 for x.. like you've got f(x) = (1+x) and when x = 0, f(0) = (1+0) = 1. So f'(0) would = 0. Can anybody help me rationalize why this is wrong? thanks in advance
The derivative measures how much f(x) changes when x changes. That is called the slope. A (suitably smooth) curved graph has a tangent line at every point, whose slope is the derivative of the function at that value of x.
Now, the function f(x) = 1+x has a graph which is a straight line. The slope never changes. So, the graph of the derivative is just the line y=1.
So, yes, when x=238423904, f'(x) is still just 1. You can't plug in different values for x, because f'(x) does not depend on x in any way.
ahh thank you so much
I can help explain why the answer provided by the person seems to contradict your expectation.
The derivative of a function measures how the function's value changes as its input (in this case, x) changes. In other words, it tells you the rate at which the function is changing at any given point.
In the case of the function f(x) = x + 1, its derivative f'(x) will always be 1. This means that at any point x, the function is changing at a constant rate of 1. Therefore, the value of x does not affect the value of the derivative.
To understand why your approach is incorrect, let's go over it step by step. You correctly identified that f(x) = 1 + x. When you plug in x = 0 into this equation, you get f(0) = 1 + 0 = 1, which is the value of the function at x = 0.
However, to find the derivative, you need to differentiate the function with respect to x. In this case, differentiating f(x) = 1 + x with respect to x gives you f'(x) = 1.
The "x" in f(x) and f'(x) represents a variable, meaning that its value can change. When you evaluate f'(x) at a specific value of x, such as x = 0, you get f'(0) = 1, which means that the rate at which the function is changing at x = 0 is 1.
It's important to note that f(x) and f'(x) are different functions. f(x) represents the original function, while f'(x) represents its derivative. The value of the derivative at a specific point, such as f'(0), tells you the rate of change of the function at that point.
In summary, the derivative of f(x) = 1 + x is constant and equal to 1, regardless of the value of x. Therefore, f'(x) = 1 for all x.