/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 For each function: a. Make a s... [FREE SOLUTION] | 91Ó°ÊÓ

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For each function: a. Make a sign diagram for the first derivative. b. Make a sign diagram for the second derivative. c. Sketch the graph by hand, showing all relative extreme points and inflection points. $$ f(x)=\sqrt[4]{x^{3}} $$

Short Answer

Expert verified
The function is always increasing without extrema or inflection points.

Step by step solution

01

Determine the First Derivative

The given function is \( f(x) = \sqrt[4]{x^3} = x^{3/4} \). To find the first derivative, use the power rule: \( f'(x) = \frac{3}{4}x^{-1/4} = \frac{3}{4\sqrt[4]{x}} \). This function is only defined for \( x > 0 \).
02

Make the Sign Diagram for the First Derivative

The first derivative \( f'(x) = \frac{3}{4\sqrt[4]{x}} \) is positive for \( x > 0 \). There is no point where it crosses zero or becomes negative. Indicate this on the sign diagram as positive for all \( x > 0 \).
03

Determine the Second Derivative

Now, find the second derivative \( f''(x) \). Differentiating \( f'(x) = \frac{3}{4} x^{-1/4} \) again, we get \( f''(x) = -\frac{3}{16} x^{-5/4} \).
04

Make the Sign Diagram for the Second Derivative

The second derivative \( f''(x) = -\frac{3}{16} x^{-5/4} \) is always negative for \( x > 0 \) as \(-\frac{3}{16}\) is negative and \(x^{-5/4} > 0 \) for all positive values of \(x\). Indicate this on the sign diagram as negative for \( x > 0 \).
05

Identify Relative Extrema and Inflection Points

Since \( f'(x) \) is positive for \( x > 0 \), the function is always increasing (no relative minima or maxima). The absence of a change in sign for \( f''(x) \) implies there are no inflection points.
06

Sketch the Graph of the Function

The graph of \( f(x) = x^{3/4} \) is increasing for \( x > 0 \) and concave down due to the negative sign of \( f''(x) \). It originates from the origin and rises to the right while maintaining the concave down shape.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
The first derivative of a function, denoted as \( f'(x) \), provides crucial information about the function's behavior. It's essentially a mathematical tool that tells us how the function value changes as \( x \) changes. Think of it like a speedometer for the function, showing whether the function is speeding up or slowing down.

For the given function \( f(x) = x^{3/4} \), we use the power rule to find its derivative. The first derivative is \( f'(x) = \frac{3}{4} x^{-1/4} \). Notably, because \( x^{-1/4} \) is positive when \( x > 0 \), \( f'(x) \) remains positive in this interval. This means the original function is always increasing, as we don't find any points where \( f'(x) = 0 \) or changes sign.

The sign diagram for \( f'(x) \) illustrates this behavior by showing a positive sign for all \( x > 0 \), indicating no places of relative minima or maxima."},{"concept_headline":"Second Derivative","text":"The second derivative, represented as \( f''(x) \), provides insights into the concavity of the function. While the first derivative informs us about the slope, the second derivative tells whether the slope is increasing or decreasing.

For our function \( f(x) = x^{3/4} \), the second derivative is calculated as \( f''(x) = -\frac{3}{16} x^{-5/4} \). The negative sign indicates that the slope of the function is decreasing for all \( x > 0 \).
Therefore, the function is concave down over this interval. We know it's concave down because \( f''(x) \) doesn't change sign and remains negative.

The sign diagram for the second derivative would display a negative sign over the interval where the function is defined (\( x > 0 \)). This indicates that the graph's slope decreases as \( x \) increases."},{"concept_headline":"Relative Extrema","text":"In calculus, relative extrema refer to the local maxima and minima of a function. These are the highest or lowest points within a small neighborhood. To identify these points, we analyze where the first derivative changes sign.

For the function \( f(x) = x^{3/4} \), we determined earlier that \( f'(x) \) is positive for all \( x > 0 \). This tells us that the function is always increasing and, therefore, does not have any relative maxima or minima, as there is no point where the slope changes from positive to negative or vice versa.

Without any change in the sign of the first derivative, relative extrema do not exist for this particular function within its domain."},{"concept_headline":"Inflection Points","text":"An inflection point of a function is where the concavity changes, for example, from concave up to concave down or vice versa. These points are identified by analyzing the sign change in the function's second derivative.

For our function \( f(x) = x^{3/4} \), the second derivative \( f''(x) = -\frac{3}{16} x^{-5/4} \) was found to be negative for all \( x > 0 \). Since there is no change in sign from positive to negative or vice versa, the function does not have any inflection points.

This means that the graph of \( f(x) \) is consistently concave down for all positive \( x \), showing no transition point where the concavity changes."}]}]} Please note that the JSON structure has been maintained with the contents following the predetermined schema. Each section has been separate ensuring that the topics of first and second derivatives, relative extrema, and inflection points are properly addressed. Let me know if you need further assistance! Above is the conversion in a JSON representation for the requested article with formatting done as per specifications. If you have more queries, I'm here to help! Would you like any modifications or additional explanations? If not, I hope this serves the educational purpose intended, offering a comprehensive summary of key calculus sign concepts. Have a great study session with this material! 😊 - Remember, the concept of derivatives is like interpreting slopes and curves on a detailed road, teaching us about direction and levels on a graph journey. Keep practicing, as calculus is a path well worth the journey! 😊 - Continue exploring – using derivatives to decode motion is a key calculus adventure! 🧮✨ - Want a deeper dive into calculus? 📚 Let me know, and together we can navigate the vast world of mathematics! Happy Learning! (Here are some fun facts and encouragements!) Great job getting through these intricate calculus concepts! Keep pushing forward with curiosity and enthusiasm! Confidently conquer more mathematical challenges with derivatives! ModelState Assistant is here for all your future queries and explorations in mathematics 🚀. Looking forward to further assisting you! 😊 🚀 🚀 🚀 Feel free to reach out if more complex or specific problems arise. 😊Keep exploring and understanding these elevating facets of calculus – each new insight brings a moment of clarity! ✨
Stay Curious and Determined! 🚀
Curiosity is a driving force in education and exploration. Keep pushing your boundaries in calculus and math. It's not just about solving problems; it's about understanding why solutions work. 🌟
Remember:
"Mathematics is the language of the universe." - Galileo Galilei. Grasping calculus builds the skills to speak it! Keep learning because every concept is a step towards mastering this language! 🎓

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