Find the maximum and minimum values of the function: f(x)

Find the maximum and minimum values of the function: $$ f(x) = \int_0^x \frac{2t+1}{t^2-2t+2} \, dt \quad \text{for} \quad x \in [1, \sqrt{3}+1] $$

To find the extreme values of $f(x)$ on the closed interval $[1, \sqrt{3}+1]$, we will:

1. Use the Fundamental Theorem of Calculus to find $f'(x)$
2. Find critical points where $f'(x) = 0$ or $f'(x)$ is undefined
3. Evaluate $f(x)$ at critical points and endpoints
4. Compare values to determine maximum and minimum

Step 1: Find the derivative using FTC

By the Fundamental Theorem of Calculus:

$$f'(x) = \frac{d}{dx}\left(\int_0^x \frac{2t+1}{t^2-2t+2} \, dt\right) = \frac{2x+1}{x^2-2x+2}$$

Step 2: Analyze the derivative

Denominator analysis:

Consider $x^2 - 2x + 2$. Completing the square:

$$x^2 - 2x + 2 = (x-1)^2 + 1$$

Since $(x-1)^2 \geq 0$ for all $x$, we have:

$$x^2 - 2x + 2 = (x-1)^2 + 1 \geq 1 > 0$$

Therefore, the denominator is always positive and never zero.

Numerator analysis:

The numerator is $2x + 1$, which equals zero when:

$$2x + 1 = 0 \implies x = -\frac{1}{2}$$

Sign of $f'(x)$:

For $x \in [1, \sqrt{3}+1]$:

• Since $x \geq 1$, we have $2x + 1 \geq 2(1) + 1 = 3 > 0$
• The denominator is always positive
• Therefore, $f'(x) > 0$ for all $x \in [1, \sqrt{3}+1]$

Step 3: Conclusion about monotonicity

Since $f'(x) > 0$ for all $x \in [1, \sqrt{3}+1]$, the function $f(x)$ is strictly increasing on this interval.

Therefore:

• The minimum value occurs at the left endpoint $x = 1$
• The maximum value occurs at the right endpoint $x = \sqrt{3}+1$

Step 4: Evaluate $f(x)$ at the endpoints

At $x = 1$:

$$f(1) = \int_0^1 \frac{2t+1}{t^2-2t+2} \, dt$$

At $x = \sqrt{3}+1$:

$$f(\sqrt{3}+1) = \int_0^{\sqrt{3}+1} \frac{2t+1}{t^2-2t+2} \, dt$$

Step 5: Evaluate the integral

To find these values, we need to compute the indefinite integral. Let's analyze:

$$\int \frac{2t+1}{t^2-2t+2} \, dt$$

Notice that the derivative of the denominator is:

$$\frac{d}{dt}(t^2-2t+2) = 2t - 2$$

We can rewrite the numerator:

$$2t + 1 = (2t - 2) + 3$$

Therefore:

$$\frac{2t+1}{t^2-2t+2} = \frac{2t-2}{t^2-2t+2} + \frac{3}{t^2-2t+2}$$

First term:

$$\int \frac{2t-2}{t^2-2t+2} \, dt = \ln|t^2-2t+2| + C_1 = \ln(t^2-2t+2) + C_1$$

(The absolute value is not needed since $t^2-2t+2 > 0$ always.)

Second term:

Since $t^2-2t+2 = (t-1)^2 + 1$, we have:

$$\int \frac{3}{(t-1)^2+1} \, dt$$

Using the substitution $u = t-1$, $du = dt$:

$$\int \frac{3}{u^2+1} \, du = 3\arctan(u) + C_2 = 3\arctan(t-1) + C_2$$

Complete antiderivative:

$$\int \frac{2t+1}{t^2-2t+2} \, dt = \ln(t^2-2t+2) + 3\arctan(t-1) + C$$

Step 6: Calculate $f(1)$

$$f(1) = \left[\ln(t^2-2t+2) + 3\arctan(t-1)\right]_0^1$$

At $t = 1$:

$$\ln(1^2-2(1)+2) + 3\arctan(1-1) = \ln(1) + 3\arctan(0) = 0 + 0 = 0$$

At $t = 0$:

$$\ln(0^2-2(0)+2) + 3\arctan(0-1) = \ln(2) + 3\arctan(-1) = \ln(2) - 3 \cdot \frac{\pi}{4} = \ln(2) - \frac{3\pi}{4}$$

Therefore:

$$f(1) = 0 - \left(\ln(2) - \frac{3\pi}{4}\right) = \frac{3\pi}{4} - \ln(2)$$

Step 7: Calculate $f(\sqrt{3}+1)$

$$f(\sqrt{3}+1) = \left[\ln(t^2-2t+2) + 3\arctan(t-1)\right]_0^{\sqrt{3}+1}$$

At $t = \sqrt{3}+1$:

\begin{align*} t^2-2t+2 &= (\sqrt{3}+1)^2 - 2(\sqrt{3}+1) + 2\\ &= 3 + 2\sqrt{3} + 1 - 2\sqrt{3} - 2 + 2\\ &= 4 \end{align*}

$$\ln(4) + 3\arctan(\sqrt{3}+1-1) = \ln(4) + 3\arctan(\sqrt{3}) = \ln(4) + 3 \cdot \frac{\pi}{3} = \ln(4) + \pi$$

At $t = 0$ (as calculated before):

$$\ln(2) - \frac{3\pi}{4}$$

Therefore:

$$f(\sqrt{3}+1) = \ln(4) + \pi - \ln(2) + \frac{3\pi}{4} = \ln\left(\frac{4}{2}\right) + \pi + \frac{3\pi}{4} = \ln(2) + \frac{7\pi}{4}$$

Final Answer

Since $f(x)$ is strictly increasing on $[1, \sqrt{3}+1]$:

Minimum value:

$$\boxed{f_{\min} = f(1) = \frac{3\pi}{4} - \ln(2) \approx 1.664}$$

Maximum value:

$$\boxed{f_{\max} = f(\sqrt{3}+1) = \ln(2) + \frac{7\pi}{4} \approx 6.201}$$

Geometric interpretation

The function $f(x) = \int_0^x \frac{2t+1}{t^2-2t+2} \, dt$ represents the signed area under the curve $y = \frac{2t+1}{t^2-2t+2}$ from $t = 0$ to $t = x$.

Key observations:

1. The integrand $g(t) = \frac{2t+1}{t^2-2t+2}$ is always positive for $t \geq 0$ since: \begin{itemize}
2. For $t \geq 0$: $2t + 1 \geq 1 > 0$
3. For all $t$: $t^2 - 2t + 2 = (t-1)^2 + 1 > 0$ \end{itemize}
4. Since the integrand is positive, $f(x)$ represents actual (unsigned) area, and this area is accumulating as $x$ increases.
5. The derivative $f'(x) = \frac{2x+1}{x^2-2x+2} > 0$ on $[1, \sqrt{3}+1]$ means the function is adding area at a positive rate throughout the interval.
6. The function increases from $f(1) \approx 1.664$ to $f(\sqrt{3}+1) \approx 6.201$, representing the accumulated area growing by approximately $4.537$ square units.

Physical interpretation: If we think of $g(t)$ as a rate (e.g., velocity), then $f(x)$ represents the total accumulation (e.g., distance traveled) from time $0$ to time $x$. Since the rate is always positive and increasing on our interval, the total accumulation is monotonically increasing.

Summary

• The function $f(x)$ has no critical points in the interior of $[1, \sqrt{3}+1]$
• It is strictly increasing on the entire interval
• Extrema occur only at the endpoints
• This is a common pattern: when $f'(x) > 0$ throughout an interval, min is at left endpoint and max is at right endpoint