Let x = -1 and x = 2 be the critical points of the function f(x) = x³ + ax² + b log_e |x| + 1, x ≠ 0

Q: Wait, why do we ignore the critical point $x = 2$?

That is a great observation! The problem asks for the absolute extrema on the specific closed interval $[-2, -0.5]$. Since $x = 2$ lies outside this interval, it does not influence the behavior of the function within our region of interest. We only care about critical points that fall inside the interval.

Q: How do I differentiate $\ln|x|$? Is it just $\frac{1}{x}$?

Yes, exactly! Whether $x$ is positive or negative, the derivative of $\ln|x|$ is $\frac{1}{x}$. This is because for $x > 0$, $\frac{d}{dx}(\ln x) = \frac{1}{x}$, and for $x < 0$, $\frac{d}{dx}(\ln(-x)) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$. So, the formula holds universally for all $x \neq 0$.

Q: Why do we need to check the endpoints $x = -2$ and $x = -0.5$?

According to the Extreme Value Theorem, the absolute maximum and minimum of a continuous function on a closed interval $[a, b]$ must occur either at a critical point within the interval or at the endpoints $a$ or $b$. If you only check the critical point, you might miss the true absolute maximum or minimum located at the boundaries.

Question

Let $x = -1$ and $x = 2$ be the critical points of the function $f(x) = x^3 + ax^2 + b \log_e |x| + 1, x 
eq 0$. Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $[-2, -1/2]$. Then $|M + m|$ is equal to (Take $\log_e 2 = 0.7$):

Options:

(A) 21.1

(B) 19.8

(C) 22.1

(D) 20.9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the function $f(x) = x^3 + ax^2 + b \ln|x| + 1$. We are given that the function has critical points at $x = -1$ and $x = 2$.

Our objective is to determine the absolute extrema of this function within the closed interval $[-2, -0.5]$.

### Decoding the Critical Points

A critical point occurs where the derivative $f'(x)$ vanishes. We differentiate $f(x)$ with respect to $x$:

$$f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(ax^2) + \frac{d}{dx}(b \ln|x|) + \frac{d}{dx}(1)$$

$$f'(x) = 3x^2 + 2ax + \frac{b}{x}$$

Since $f'(x) = 0$ at $x = -1$ and $x = 2$, we can establish a system of equations to solve for the constants $a$ and $b$.

### The Algebraic Dance

At $x = -1$:

$$3(-1)^2 + 2a(-1) + \frac{b}{-1} = 0 \Rightarrow 3 - 2a - b = 0 \Rightarrow 2a + b = 3$$

At $x = 2$:

$$3(2)^2 + 2a(2) + \frac{b}{2} = 0 \Rightarrow 12 + 4a + \frac{b}{2} = 0$$

Multiplying the second equation by $2$ to clear the fraction, we obtain $24 + 8a + b = 0$, or $8a + b = -24$.

Subtracting the first equation ($2a + b = 3$) from the second ($8a + b = -24$):

$$6a = -27 \Rightarrow a = -4.5$$

Substituting $a = -4.5$ into $2a + b = 3$:

$$2(-4.5) + b = 3 \Rightarrow -9 + b = 3 \Rightarrow b = 12$$

The function is thus defined as $f(x) = x^3 - 4.5x^2 + 12 \ln|x| + 1$.

### The Extreme Value Hunt

To find the absolute extrema on $[-2, -0.5]$, we evaluate $f(x)$ at the critical point $x = -1$ (which lies within the interval) and at the endpoints $x = -2$ and $x = -0.5$.

Evaluating at $x = -1$:

$$f(-1) = (-1)^3 - 4.5(-1)^2 + 12 \ln|-1| + 1 = -1 - 4.5 + 0 + 1 = -4.5$$

Evaluating at $x = -2$:

$$f(-2) = (-2)^3 - 4.5(-2)^2 + 12 \ln|-2| + 1 = -8 - 18 + 12 \ln(2) + 1$$

Using $\ln(2) \approx 0.693$, we get $f(-2) \approx -25 + 8.316 = -16.684$.

Evaluating at $x = -0.5$:

$$f(-0.5) = (-0.5)^3 - 4.5(-0.5)^2 + 12 \ln|-0.5| + 1 = -0.125 - 1.125 + 12(-\ln 2) + 1$$

$$f(-0.5) = -0.25 - 12(0.693) \approx -0.25 - 8.316 = -8.566$$

### Final Calculation

Comparing the values $\{-4.5, -16.684, -8.566\}$, we identify:

**Absolute Maximum ($M$)** $= -4.5$

**Absolute Minimum ($m$)** $\approx -16.684$

The final result is $|M + m| = |-4.5 - 16.684| = \mathbf{21.184}$.

Select Answer:

Visualized Solution

Function and Critical Points

Derivative $f^{'} (x)$

First Equation

Second Equation

Solving for $a$ and $b$

The Complete Function

Testing $x = - 1$

Testing $x = - 2$

Testing $x = - 0.5$

Identifying $M$ and $m$

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Critical Points

The Algebraic Dance

The Extreme Value Hunt

Final Calculation

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Select Answer:

Visualized Solution

Function and Critical Points

Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

First Equation

Second Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Complete Function

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−2

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−0.5

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Critical Points

The Algebraic Dance

The Extreme Value Hunt

Final Calculation

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