Find f(x) so that it satisfies:
f'(x)=1+e^x+1/x and f(1)=3+e. Another tutor was kind enough to give me the anti deriv as x+e^x+ln x+c and that the constant is 2. But I can't figure how to prove it or where to start so I will know for the test.
Thanks.
To find f(x) such that f'(x) = 1 + e^x + 1/x and f(1) = 3 + e, we need to integrate the given expression with respect to x to get the antiderivative of f'(x).
Starting with the given expression f'(x) = 1 + e^x + 1/x, we can integrate each term separately.
The integral of 1, with respect to x, is x.
The integral of e^x, with respect to x, is e^x.
To find the integral of 1/x, we can use the natural logarithm function. The integral of 1/x, with respect to x, is ln|x|.
Therefore, the antiderivative of f'(x) = 1 + e^x + 1/x is f(x) = x + e^x + ln|x| + C, where C is the constant of integration.
Next, we have to determine the value of the constant C. We are given that f(1) = 3 + e.
Plugging x = 1 into the equation f(x), we get:
f(1) = 1 + e^1 + ln|1| + C
f(1) = 1 + e + ln(1) + C
f(1) = 1 + e + 0 + C
f(1) = 1 + e + C
We also know that f(1) is equal to 3 + e. Therefore, we can equate these two expressions:
3 + e = 1 + e + C
Simplifying the equation, we have:
2 = C
So the constant of integration, C, is equal to 2.
Finally, we can rewrite the expression for f(x) using the determined value of C:
f(x) = x + e^x + ln|x| + 2
Therefore, the function f(x) that satisfies f'(x) = 1 + e^x + 1/x and f(1) = 3 + e is f(x) = x + e^x + ln|x| + 2.