Let Max{0 ≤ x ≤ 2} {(9-x²)/(5-x)} = α and Min{0 ≤ x ≤ 2} {(9-x²)/(5-x)} = β. If ∫[β - 8/3 to 2α - 1]

Q: Wait, why do we reject the root $x=9$ when finding critical points?

That is a great question! When we solve $f'(x) = 0$, we find the points where the slope of the function is zero. Our derivative calculation yielded $x=1$ and $x=9$. However, the problem explicitly restricts our domain to the interval $x \in [0, 2]$. Since $9$ is outside this interval, it cannot be a local maximum or minimum within our range of interest. Always check your domain constraints!

Q: Why do we need to find the intersection point $x=1.8$?

The integral is of the function $\text{Max} \{ f(x), x \}$. This means the integrand is not a single function, but rather the 'upper envelope' of two different curves. At $x=1.8$, the two curves $f(x)$ and $y=x$ cross. Before $1.8$, one is larger; after $1.8$, the other is larger. To integrate this, we must split the integral at the intersection point to correctly identify which function is the 'maximum' in each sub-interval.

Q: Is there a trick to integrating $\frac{9-x^2}{5-x}$?

It looks intimidating, but the trick is polynomial long division. By rewriting $\frac{9-x^2}{5-x}$ as $x + 5 + \frac{16}{x-5}$, you turn a complex rational function into a simple polynomial plus a logarithmic term. It makes the integration process much cleaner and less prone to errors.

Question

Let $	ext{Max}_{0 \le x \le 2} \left\{ \frac{9-x^2}{5-x} ight\} = \alpha$ and $	ext{Min}_{0 \le x \le 2} \left\{ \frac{9-x^2}{5-x} ight\} = \beta$. If $\int_{\beta - \frac{8}{3}}^{2\alpha - 1} 	ext{Max} \left\{ \frac{9-x^2}{5-x}, x ight\} dx = \alpha_1 + \alpha_2 \log_e \left( \frac{8}{15} ight)$, then $\alpha_1 + \alpha_2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the function $f(x) = \frac{9-x^2}{5-x}$ defined on the interval $[0, 2]$. To find the absolute maximum $\alpha$ and minimum $\beta$, we apply the quotient rule to find the derivative:

$$f'(x) = \frac{(5-x)(-2x) - (9-x^2)(-1)}{(5-x)^2} = \frac{-10x + 2x^2 + 9 - x^2}{(5-x)^2} = \frac{x^2 - 10x + 9}{(5-x)^2}$$

Setting $f'(x) = 0$ yields the critical points $x=1$ and $x=9$. Since we are restricted to the interval $[0, 2]$, we discard $x=9$ as it lies outside our domain.

### Determining Extrema

We evaluate the function at the critical point $x=1$ and the boundaries $x=0$ and $x=2$:

1. $f(0) = \frac{9-0}{5-0} = 1.8$
2. $f(1) = \frac{9-1}{5-1} = \frac{8}{4} = 2$
3. $f(2) = \frac{9-4}{5-2} = \frac{5}{3}$

Comparing these values, we identify the absolute maximum $\alpha = 2$ and the absolute minimum $\beta = \frac{5}{3}$.

### The Geometry of the Integral

We now evaluate the integral $I = \int_{\beta - \frac{8}{3}}^{2\alpha - 1} 	ext{Max} \{ f(x), x \} dx$. Substituting $\alpha = 2$ and $\beta = \frac{5}{3}$, the limits of integration become:

Lower limit: $\frac{5}{3} - \frac{8}{3} = -1$

Upper limit: $2(2) - 1 = 3$

To determine the "upper envelope" $	ext{Max} \{ f(x), x \}$, we find the intersection of $f(x)$ and $y=x$:

$$\frac{9-x^2}{5-x} = x \Rightarrow 9-x^2 = 5x - x^2 \Rightarrow 5x = 9 \Rightarrow x = 1.8$$

For $x < 1.8$, $f(x)$ is the dominant function, while for $x > 1.8$, the line $y=x$ is greater.

### The Algebraic Grind

We split the integral at the pivot point $x=1.8$:

$$I = \int_{-1}^{1.8} \left( \frac{9-x^2}{5-x} ight) dx + \int_{1.8}^{3} x dx$$

Using polynomial long division, we rewrite the integrand as $f(x) = x + 5 + \frac{16}{x-5}$. The integration proceeds as follows:

$$\int (x + 5 + \frac{16}{x-5}) dx = \frac{x^2}{2} + 5x + 16 \ln|x-5|$$

Evaluating the first part from $-1$ to $1.8$:

$$\left[ \frac{1.8^2}{2} + 5(1.8) + 16 \ln|1.8-5| ight] - \left[ \frac{(-1)^2}{2} + 5(-1) + 16 \ln|-1-5| ight] = 15.12 + 16 \ln(3.2) - (-4.5 + 16 \ln 6) = 19.62 + 16 \ln(\frac{3.2}{6}) = 19.62 + 16 \ln(\frac{8}{15})$$

### Final Calculation

The second part of the integral is:

$$\int_{1.8}^{3} x dx = \left[ \frac{x^2}{2} ight]_{1.8}^{3} = 4.5 - 1.62 = 2.88$$

Summing the two parts:

$$I = 19.62 + 2.88 + 16 \ln(\frac{8}{15}) = 22.5 + 16 \ln(\frac{8}{15})$$

*(Note: Based on the provided logic flow, $\alpha_1 = 22.5$ and $\alpha_2 = 16$)*. The final result is **38.5**.

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Find Critical Points

Determine $α$ and $β$

Set Up the Integral

Find Intersection Point

Identify the Max Function

Split the Integral

Simplify the First Integrand

Integrate the First Part

Evaluate Limits for First Part

Integrate the Second Part

Final Result and Comparison

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Determining Extrema

The Geometry of the Integral

The Algebraic Grind

Final Calculation

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