first picture find Y if at the point (a,f(a))(a,f(a)) for a=2second picture Find h′(1)h′(1) for the function h(x)=f(x)f(x). please and thank you 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 =

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Answered: Let f be function such that f(1) = 4…

Let f be function such that f(1) = 4 and ƒ'(1) = 11.

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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