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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Question
first picture find Y if at the point (a,f(a))(a,f(a)) for a=2
second picture Find h′(1)h′(1) for the function h(x)=f(x)f(x).
please and thank you
Transcribed Image Text:Let \( f \) be a function such that \( f(1) = 4 \) and \( f'(1) = 11 \).
1) For the function \( h(x) = f(x)^2 \), find \( h'(1) \).
![**Problem Statement:**
Find an equation for the line tangent to the graph of
\[ f(x) = 7xe^x \]
at the point \((a, f(a))\) for \(a = 2\).
**Solution:**
To find the equation of the tangent line to the curve at a given point, follow these steps:
1. **Differentiate \(f(x)\):**
Find the derivative \(f'(x)\) of the function \(f(x) = 7xe^x\). Use the product rule since the function is a product of \(7x\) and \(e^x\).
\[ f'(x) = \frac{d}{dx}[7x \cdot e^x] = 7 \cdot (x \cdot e^x)' = 7 \cdot (1 \cdot e^x + x \cdot e^x) = 7(e^x + xe^x) \]
Therefore,
\[ f'(x) = 7e^x(1 + x) \]
2. **Evaluate the derivative at \(a = 2\):**
Calculate \(f'(2)\) to find the slope of the tangent line at \(x = 2\).
\[ f'(2) = 7e^2(1 + 2) = 21e^2 \]
3. **Calculate \(f(2)\):**
Find the y-coordinate of the point on the curve by evaluating the original function at \(x = 2\).
\[ f(2) = 7 \times 2 \times e^2 = 14e^2 \]
4. **Write the equation of the tangent line:**
Use the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency, \((2, 14e^2)\).
\[ y - 14e^2 = 21e^2(x - 2) \]
Simplifying the expression gives:
\[ y = 21e^2x - 28e^2 + 14e^2 \]
\[ y =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb11ca76e-99cc-49cc-ab47-e18e3e84e4f7%2F366bfc93-0e21-4569-8f5f-425718e8cc9f%2Frmks8vv_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation for the line tangent to the graph of
\[ f(x) = 7xe^x \]
at the point \((a, f(a))\) for \(a = 2\).
**Solution:**
To find the equation of the tangent line to the curve at a given point, follow these steps:
1. **Differentiate \(f(x)\):**
Find the derivative \(f'(x)\) of the function \(f(x) = 7xe^x\). Use the product rule since the function is a product of \(7x\) and \(e^x\).
\[ f'(x) = \frac{d}{dx}[7x \cdot e^x] = 7 \cdot (x \cdot e^x)' = 7 \cdot (1 \cdot e^x + x \cdot e^x) = 7(e^x + xe^x) \]
Therefore,
\[ f'(x) = 7e^x(1 + x) \]
2. **Evaluate the derivative at \(a = 2\):**
Calculate \(f'(2)\) to find the slope of the tangent line at \(x = 2\).
\[ f'(2) = 7e^2(1 + 2) = 21e^2 \]
3. **Calculate \(f(2)\):**
Find the y-coordinate of the point on the curve by evaluating the original function at \(x = 2\).
\[ f(2) = 7 \times 2 \times e^2 = 14e^2 \]
4. **Write the equation of the tangent line:**
Use the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency, \((2, 14e^2)\).
\[ y - 14e^2 = 21e^2(x - 2) \]
Simplifying the expression gives:
\[ y = 21e^2x - 28e^2 + 14e^2 \]
\[ y =
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