f(x) is a function differentiable at x=1 and f�Œ(1)=1/8. What is the value of f(x)-f(1)
f'(x) denotes the derivative of f(x)
To find the value of f(x) - f(1), you can use the Mean Value Theorem for Differentiable Functions. According to the Mean Value Theorem, if f(x) is a function differentiable on an interval [a, b], then there exists at least one point c in the interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, since f(x) is differentiable at x = 1, we can use the Mean Value Theorem to find the value of f(x) - f(1). Let's denote a = 1 and b = x. Using the Mean Value Theorem, we have:
f'(c) = (f(x) - f(1))/(x - 1)
Now, we know that f'(x) denotes the derivative of f(x). So, in this case, f'(c) represents the derivative of f(x) at point c.
Since f(x) is differentiable at x = 1, we can use the given information f�Œ(1)=1/8, which means f'(1) = 1/8.
Substituting the values into the equation from the Mean Value Theorem, we get:
1/8 = (f(x) - f(1))/(x - 1)
To find the value of f(x) - f(1), we need to isolate it. Multiply both sides of the equation by (x - 1):
1/8 * (x - 1) = f(x) - f(1)
Simplifying further, we have:
(x - 1)/8 = f(x) - f(1)
So, the value of f(x) - f(1) is (x - 1)/8.