Can you help me with this problem with step by step explanation, please? Thank you:) ### Finding Points of Horizontal Tangency**Problem Statement:**Use a graphing utility to graph $f$ and $f'$ over the given interval. Determine any points at which the graph of $f$ has horizontal tangents. (Order your answers from smallest to largest $x$, then from smallest to largest $y$. Round your answers to three decimal places.)**Given Functions and Interval:**\[ f(x) = x^3 - 1.3x^2 - 0.97x + 1.41 \]Interval: $[-2, 2]$**Solution:**The potential points of horizontal tangency are evaluated, and their coordinates are shown below:1. $(x, y) = \left( -0.282, 0 \right)$ - Incorrect2. $(x, y) = \left( 1.148, 0 \right)$ - Incorrect**Explanation of Diagram and Results:**The image includes the following elements:- A given polynomial function $f(x)$.- An instruction to use graphing utility software to plot the function and its derivative over a specified interval.- Two evaluated points $(-0.282, 0)$ and $ (1.148, 0)$, each marked with a red cross (×), indicating that these points are incorrect for horizontal tangency.To accurately determine the points of horizontal tangency:- The derivative $f'(x)$ should be computed.- The roots of $f'(x)$ (the $x$-values where $f'(x) = 0$) are found, ensuring all potential horizontal tangent points within the given interval are obtained.

Answered: Image /qna-images/answer/380f8a61-3999-4e1f-9567-2af64e4c5bbc.jpg

Use a graphing utility to graph f and f'over the given interval. Determine any points at which the graph of f has horizontal tangents. (Order your answers from smallest to largest x, then from smallest to largest y. Round your answers to three decimal places. (x) - x - 1.3x? - 0.97x + 1.41; (-2, 2) (х, у)- -0.282, 0 (х, у) - 1.148, 0

Use a graphing utility to graph f and f'over the given interval. Determine any points at which the graph of f has horizontal tangents. (Order your answers from smallest to largest x, then from smallest to largest y. Round your answers to three decimal places. (x) - x - 1.3x? - 0.97x + 1.41; (-2, 2) (х, у)- -0.282, 0 (х, у) - 1.148, 0

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Can you help me with this problem with step by step explanation, please? Thank you:)

### Finding Points of Horizontal Tangency

**Problem Statement:**
Use a graphing utility to graph $f$ and $f'$ over the given interval. Determine any points at which the graph of $f$ has horizontal tangents. (Order your answers from smallest to largest $x$, then from smallest to largest $y$. Round your answers to three decimal places.)

**Given Functions and Interval:**
\[ f(x) = x^3 - 1.3x^2 - 0.97x + 1.41 \]
Interval: $[-2, 2]$

**Solution:**
The potential points of horizontal tangency are evaluated, and their coordinates are shown below:

1. $(x, y) = \left( -0.282, 0 \right)$ - Incorrect
2. $(x, y) = \left( 1.148, 0 \right)$ - Incorrect

**Explanation of Diagram and Results:**
The image includes the following elements:

- A given polynomial function $f(x)$.
- An instruction to use graphing utility software to plot the function and its derivative over a specified interval.
- Two evaluated points $(-0.282, 0)$ and $ (1.148, 0)$, each marked with a red cross (×), indicating that these points are incorrect for horizontal tangency.

To accurately determine the points of horizontal tangency:
- The derivative $f'(x)$ should be computed.
- The roots of $f'(x)$ (the $x$-values where $f'(x) = 0$) are found, ensuring all potential horizontal tangent points within the given interval are obtained.

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