Find f(x) if f(1) = 1 and the tangent line at (x, f(x)) has slope 6/x
f' = 6/x
f = 6ln(x)+c
1 = 6ln(1)+c
c=1
f(x) = 6ln(x)+1
To find the function f(x) given the given information, we need to integrate the given slope expression with respect to x.
The slope of the tangent line at any point (x, f(x)) on the graph of f(x) can be found using the derivative of f(x). Since the slope is given as 6/x, we can write this relationship between the derivative and the slope as:
f'(x) = 6/x
To find the actual function f(x) from the derivative, we need to integrate f'(x).
∫ f'(x) dx = ∫ (6/x) dx
Let's integrate the right-hand side of the equation:
∫ (6/x) dx
To integrate this, we can rewrite it as:
6∫ (1/x) dx
Now, we can integrate with respect to x:
6 ∫ (1/x) dx = 6 ln|x| + C
where ln|x| is the natural logarithm of the absolute value of x, and C is the constant of integration.
So, the function f(x) is:
f(x) = 6 ln|x| + C
Now, to find the constant of integration, C, we can use the given initial condition f(1) = 1.
Substituting x = 1 and f(x) = 1 into the equation for f(x), we get:
1 = 6 ln|1| + C
Since ln(1) = 0, the equation simplifies to:
1 = C
Therefore, the constant of integration is C = 1.
Finally, we can write the equation for f(x) as:
f(x) = 6 ln|x| + 1