RaySix

Post

at November 1st 2023, 3:21:23 pm.

Question:

Consider the function

f(x) = \int \frac{5x^2 + x + 3}{2x+1} \,dx

where $f'(x)$ is differentiable for all $x \neq -\frac{1}{2}$ .

(a) Find $f'(x)$ .

(b) Determine the equation of the line tangent to the graph of $f$ at the point where $x = 2$ .

(c) Find the value of $x$ where $f(x)$ has a relative minimum.

Answer:

(a) To find $f'(x)$ , we will use the Fundamental Theorem of Calculus. We will rewrite $f(x)$ using long division:

\int \frac{5x^2 + x + 3}{2x+1} \,dx = \int 2x + 1 + \frac{2}{2x+1} \,dx

Now applying the Fundamental Theorem of Calculus:

f(x) = x^2 + x + 2\ln|2x+1| + C

where $C$ is the constant of integration.

Therefore, $f'(x)$ is given by:

f'(x) = 2x + 1 + \frac{2}{2x+1}

(b)

f'(2) = 2(2) + 1 + \frac{2}{2(2)+1} = 5

The slope of the tangent line is 5. To find the $y$ -coordinate of the point where $x = 2$ , we substitute $x=2$ into $f(x)$ :

f(2) = (2)^2 + (2) + 2\ln|2(2)+1| + C = 10 + 2\ln 5 + C

Hence, the equation of the tangent line is:

y - f(2) = 5(x - 2)

Simplifying the equation:

y = 5x - 10 + 10 + 2\ln 5 + C

y = 5x + 2\ln 5 + C

(c)

2x + 1 + \frac{2}{2x+1} = 0

Multiplying the equation by $2x+1$ to clear the denominator:

(2x+1)(2x+1) + 2 = 0

Simplifying the equation:

4x^2 + 4x + 1 + 2 = 0

4x^2 + 4x + 3 = 0

However, this equation has no real solutions. Therefore, there is no value of $x$ where $f(x)$ has a relative minimum.