Find the maximum and minimum values of the function: y =

Find the maximum and minimum values of the function: $$ y = \int_0^x \frac{2t-1}{t^2-t-1} \, dt \quad \text{for} \quad x \in [0,1] $$

To find the extreme values of $y(x)$ on the closed interval $[0, 1]$, we will:

1. Use the Fundamental Theorem of Calculus to find $y'(x)$
2. Find critical points where $y'(x) = 0$ or $y'(x)$ is undefined
3. Evaluate $y(x)$ 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:

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

Step 2: Analyze the denominator

Consider $x^2 - x - 1 = 0$. Using the quadratic formula:

$$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$$

The roots are:

$$x_1 = \frac{1 - \sqrt{5}}{2} \approx -0.618 \quad \text{and} \quad x_2 = \frac{1 + \sqrt{5}}{2} \approx 1.618$$

Note that $x_2 = \phi$ (the golden ratio).

Important: Both roots lie outside the interval $[0, 1]$:

• $x_1 < 0$
• $x_2 > 1$

Therefore, the denominator $x^2 - x - 1$ is never zero on $[0, 1]$.

Sign of the denominator on $[0, 1]$:

Let's check at $x = 0$:

$$0^2 - 0 - 1 = -1 < 0$$

Since the parabola $x^2 - x - 1$ has no roots in $[0, 1]$ and is negative at $x = 0$, it remains negative throughout $[0, 1]$.

Step 3: Find critical points

Critical points occur where $y'(x) = 0$:

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

We need to verify that $x = \frac{1}{2} \in [0, 1]$. ✓

Let's also verify the denominator is non-zero at $x = \frac{1}{2}$:

$$\left(\frac{1}{2}\right)^2 - \frac{1}{2} - 1 = \frac{1}{4} - \frac{1}{2} - 1 = -\frac{5}{4} \neq 0$$

So $x = \frac{1}{2}$ is indeed a critical point.

Step 4: Determine the nature of the critical point

Let's analyze the sign of $y'(x)$ on $[0, 1]$.

Since the denominator $x^2 - x - 1 < 0$ for all $x \in [0, 1]$, the sign of $y'(x)$ is determined by the opposite of the sign of the numerator:

• For $x \in [0, \frac{1}{2})$: $2x - 1 < 0$, so $\frac{2x-1}{x^2-x-1} = \frac{\text{negative}}{\text{negative}} > 0$
• For $x = \frac{1}{2}$: $2x - 1 = 0$, so $y'(\frac{1}{2}) = 0$
• For $x \in (\frac{1}{2}, 1]$: $2x - 1 > 0$, so $\frac{2x-1}{x^2-x-1} = \frac{\text{positive}}{\text{negative}} < 0$

Sign chart for $y'(x)$:

$$\begin{array}{c|ccc} x & 0 & \frac{1}{2} & 1 \\ \hline y'(x) & + & 0 & - \\ y(x) & \nearrow & \text{max} & \searrow \end{array}$$

Conclusion: Since $y'(x)$ changes from positive to negative at $x = \frac{1}{2}$, this point is a local maximum. Moreover, since it's the only critical point in $[0, 1]$, it is the global maximum on this interval.

Step 5: Evaluate $y$ at critical point and endpoints

We need to evaluate $y$ at three points: $x = 0$, $x = \frac{1}{2}$, and $x = 1$.

At $x = 0$:

$$y(0) = \int_0^0 \frac{2t-1}{t^2-t-1} \, dt = 0$$

At $x = \frac{1}{2}$:

$$y\left(\frac{1}{2}\right) = \int_0^{1/2} \frac{2t-1}{t^2-t-1} \, dt$$

At $x = 1$:

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

Step 6: Evaluate the integral

To compute these values, we need the antiderivative:

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

Notice that the derivative of the denominator is:

$$\frac{d}{dt}(t^2-t-1) = 2t - 1$$

This is exactly the numerator! Therefore:

$$\int \frac{2t-1}{t^2-t-1} \, dt = \ln|t^2-t-1| + C$$

Since $t^2 - t - 1 < 0$ for $t \in [0, 1]$, we have:

$$\ln|t^2-t-1| = \ln(-(t^2-t-1)) = \ln(-t^2+t+1)$$

Step 7: Calculate $y(1/2)$

$$y\left(\frac{1}{2}\right) = \left[\ln(-t^2+t+1)\right]_0^{1/2}$$

At $t = \frac{1}{2}$:

$$-\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 1 = -\frac{1}{4} + \frac{1}{2} + 1 = \frac{5}{4}$$

At $t = 0$:

$$-0^2 + 0 + 1 = 1$$

Therefore:

$$y\left(\frac{1}{2}\right) = \ln\left(\frac{5}{4}\right) - \ln(1) = \ln\left(\frac{5}{4}\right) = \ln(5) - \ln(4)$$

Step 8: Calculate $y(1)$

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

At $t = 1$:

$$-1^2 + 1 + 1 = 1$$

At $t = 0$:

$$1$$

Therefore:

$$y(1) = \ln(1) - \ln(1) = 0$$

Step 9: Compare values

We have:

\begin{align*} y(0) &= 0\\ y\left(\frac{1}{2}\right) &= \ln\left(\frac{5}{4}\right) \approx 0.223\\ y(1) &= 0 \end{align*}

Final Answer

Maximum value:

$$\boxed{y_{\max} = y\left(\frac{1}{2}\right) = \ln\left(\frac{5}{4}\right) = \ln(5) - \ln(4) \approx 0.223}$$

Minimum value:

$$\boxed{y_{\min} = y(0) = y(1) = 0}$$

The minimum is achieved at both endpoints.

Geometric interpretation

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

Key observations:

1. The integrand $g(t) = \frac{2t-1}{t^2-t-1}$ changes sign at $t = \frac{1}{2}$: \begin{itemize}
2. For $t \in [0, \frac{1}{2})$: numerator is negative, denominator is negative, so $g(t) > 0$
3. For $t \in (\frac{1}{2}, 1]$: numerator is positive, denominator is negative, so $g(t) < 0$ \end{itemize}
4. From $t = 0$ to $t = \frac{1}{2}$, the function $g(t)$ is positive, so area is being added. This increases $y(x)$ from 0 to $\ln(5/4)$.
5. From $t = \frac{1}{2}$ to $t = 1$, the function $g(t)$ is negative, so area is being subtracted. This decreases $y(x)$ back to 0.
6. The positive area from $[0, \frac{1}{2}]$ exactly cancels the negative area from $[\frac{1}{2}, 1]$, which is why $y(1) = 0$.
7. The maximum accumulated area occurs at the transition point $x = \frac{1}{2}$, where the integrand changes from positive to negative.

Visual description: Imagine filling a container with water at $t \in [0, \frac{1}{2}]$ (positive rate) and then draining it at $t \in [\frac{1}{2}, 1]$ (negative rate). The maximum water level occurs at $t = \frac{1}{2}$, and by $t = 1$, the container is empty again.

Summary

• The function has one critical point at $x = \frac{1}{2}$ in the interior of $[0, 1]$
• The function increases on $[0, \frac{1}{2}]$ and decreases on $[\frac{1}{2}, 1]$
• The maximum occurs at the critical point $x = \frac{1}{2}$
• The minimum (value 0) occurs at both endpoints
• This illustrates that extrema can occur at critical points (interior) or at endpoints
• The integral from 0 to 1 equals zero because the positive and negative areas cancel

Key lesson: When the integrand changes sign, the accumulated area function can increase and then decrease, leading to interior extrema. Always check critical points in addition to endpoints!