The sum of absolute maximum and absolute minimum values of the function f(x) = |2x² + 3x - 2| + sin

Q: Wait, why can't we just differentiate the modulus function directly?

That is a classic trap! The modulus function $|g(x)|$ is not differentiable at the points where $g(x) = 0$. If you try to differentiate it directly without splitting it into cases, you will miss the behavior at the critical point $x = 1/2$. Always define the function piecewise first to ensure your calculus is rigorous.

Q: How do I know for sure whether the maximum is at $x = 0$ or $x = 1$?

In problems like this, the derivative test tells us the function is monotonic on either side of the critical point. Since it decreases until $x = 1/2$ and increases thereafter, the global maximum must be at one of the boundaries of the interval. Simply evaluate $f(0)$ and $f(1)$ and compare the values—it is the most reliable way to be certain.

Q: Is the derivative test always necessary, or can I just plug in points?

Plugging in points is a good way to check your work, but it is not a proof. The derivative test is your mathematical 'safety net.' It confirms the function's behavior across the entire interval, ensuring you haven't missed any hidden local extrema or inflection points that might change the global maximum or minimum.

Question

The sum of absolute maximum and absolute minimum values of the function $f(x) = |2x^2 + 3x - 2| + \sin x \cos x$ in the interval $[0, 1]$ is :

Options:

(A) $3 + \frac{\sin(1) \cos^2(1/2)}{2}$

(B) $3 + \frac{1}{2}(1 + 2\cos(1)) \sin(1)$

(C) $5 + \frac{1}{2}(\sin(1) + \sin(2))$

(D) $2 + \sin(\frac{1}{2}) \cos(\frac{1}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup The function is defined as $f(x) = |2x^2 + 3x - 2| + \sin x \cos x$ on the interval $[0, 1]$. To analyze this, we must first address the absolute value component and the trigonometric product. ### Taming the Quadratic Beast The expression inside the modulus is $2x^2 + 3x - 2$. Factoring this quadratic, we obtain: $$2x^2 + 3x - 2 = (2x - 1)(x + 2)$$ The roots of this quadratic are $x = 1/2$ and $x = -2$. Within our interval of interest $[0, 1]$, the critical point is $x = 1/2$. For $x \in [0, 1/2)$, the quadratic is negative, so the modulus flips the sign. For $x \in [1/2, 1]$, the quadratic is positive, and the modulus leaves it unchanged. ### The Trigonometric Simplification The second term, $\sin x \cos x$, can be simplified using the double-angle identity $\sin 2x = 2 \sin x \cos x$. This allows us to rewrite the term as: $$\sin x \cos x = \frac{1}{2} \sin 2x$$ This substitution transforms the product into a single, smooth oscillation, making the function easier to differentiate. ### The Monotonicity Dance We now examine the derivative $f'(x)$ to determine the behavior of the function. For $x < 1/2$: $$f'(x) = -(4x + 3) + \cos 2x$$ Since $x \ge 0$, the term $-(4x + 3)$ is at most $-3$, while $\cos 2x$ is at most $1$. Thus, $f'(x) < 0$, indicating the function is strictly decreasing on $[0, 1/2)$. For $x > 1/2$: $$f'(x) = (4x + 3) + \cos 2x$$ Here, $4x + 3 > 5$, and since $\cos 2x \ge -1$, the derivative $f'(x)$ is always positive. The function is strictly increasing on $(1/2, 1]$. ### The Final Victory The absolute minimum occurs at the "valley" of $x = 1/2$, while the absolute maximum occurs at one of the endpoints, $x = 0$ or $x = 1$. We calculate the values: $$f(1/2) = |0| + \frac{1}{2} \sin 1 = \frac{1}{2} \sin 1$$ $$f(0) = |-2| + 0 = 2$$ $$f(1) = |2 + 3 - 2| + \frac{1}{2} \sin 2 = 3 + \frac{1}{2} \sin 2$$ Comparing these values, $3 + \frac{1}{2} \sin 2$ is the absolute maximum. The sum of the absolute maximum and absolute minimum is: $$3 + \frac{1}{2} \sin 2 + \frac{1}{2} \sin 1$$ Using the identity $\sin 2 = 2 \sin 1 \cos 1$, the expression becomes $3 + \sin 1 \cos 1 + \frac{1}{2} \sin 1$. The final result is: **$$3 + \frac{1}{2} \sin 1 (1 + 2 \cos 1)$$**

Animated Solution for Mathematics - Differentiation: The sum of absolute maximum and absolute minimum values of the function $f (x) = ∣2 x^{2} + 3 x - 2∣ + sin x cos x$ in the interval $[0, 1]$ is :

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Visualized Solution

Function Analysis for $f (x)$

Factoring $2 x^{2} + 3 x - 2$

Sign Analysis in $[0, 1]$

Simplifying $sin x cos x$

Piecewise Definition of $f (x)$

Derivative for $x < \frac{1}{2}$

Derivative for $x > \frac{1}{2}$

Absolute Minimum at $x = \frac{1}{2}$

Absolute Maximum at Endpoints

Final Summation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Taming the Quadratic Beast

The Trigonometric Simplification

The Monotonicity Dance

The Final Victory

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Visualized Solution

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

Factoring .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x2+3x−2

Sign Analysis in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,1]

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinxcosx

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

Derivative for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<21​

Derivative for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>21​

Absolute Minimum at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=21​