Question

In Exercises \(35-38\) , find the function with the given derivative whose graph passes through the point \(P\) . $$f^{\prime}(x)=\frac{1}{x+2}, \quad x>-2, \quad P(-1,3)$$

Short answer

The function with the given derivative that passes through the point \((-1,3)\) is \(f(x) = \ln(x+2) + 3\).

Step-by-step solution

Integral Calculation

We begin by finding the indefinite integral or antiderivative of the given derivative, \(f'(x) = \frac{1}{x + 2}\). This can be done by applying the rule of integrals. Integral of \(f'(x) = \frac{1}{x + 2}\) is \(\ln|x+2|\). But, as we're told that \(x > -2\), the absolute value isn't necessary, so the antiderivative function \(F(x)\) can be expressed as \(F(x) = \ln(x+2) + C\), where \(C\) is the constant of integration.

Calculation of Constant

We're given that the function passes through the point \((-1,3)\), which means when \(x = -1\), \(F(x) = 3\). We substitute these values into the function equation \(F(x) = \ln(x+2) + C\) to solve for C. Thus, we get \(3 = \ln((-1)+2) + C\), which simplifies to \(3 = \ln{1} + C\). Since \(\ln{1} = 0\), the constant \(C = 3\).

Formulate Complete Function

Now that we've determined that the constant of integration \(C = 3\), we can substitute this value back into our function equation to get the final function. Therefore, the function \(f(x)\) becomes \(f(x) = \ln(x+2) + 3\).