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F. Find the linear aproxuimation to the following function at fie given point a. \[ f(x)=-2 x^{2}+x a=1 \]

Ask by Cole Coleman. in the United States
Feb 03,2025

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The linear approximation of \( f(x) = -2x^2 + x \) at \( a = 1 \) is \( L(x) = -3x + 2 \).

Solution

To find the linear approximation (also known as the tangent line approximation) of the function \( f(x) = -2x^2 + x \) at the point \( a = 1 \), follow these steps: **1. Compute \( f(a) \):** \[ f(1) = -2(1)^2 + 1 = -2 + 1 = -1 \] **2. Find the derivative \( f'(x) \):** \[ f'(x) = \frac{d}{dx} (-2x^2 + x) = -4x + 1 \] **3. Compute \( f'(a) \):** \[ f'(1) = -4(1) + 1 = -4 + 1 = -3 \] **4. Write the linear approximation formula:** \[ L(x) = f(a) + f'(a)(x - a) \] **5. Substitute the values into the formula:** \[ L(x) = -1 + (-3)(x - 1) = -1 - 3x + 3 = -3x + 2 \] **Final Answer:** The linear approximation at \( a = 1 \) is \( L(x) = -3\,x + 2 \).

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To find the linear approximation of the function \( f(x) = -2x^2 + x \) at the point \( a = 1 \), we first need to find \( f(1) \) and the derivative \( f'(x) \). Calculating \( f(1) \): \[ f(1) = -2(1)^2 + (1) = -2 + 1 = -1. \] Next, we find the derivative \( f'(x) \): \[ f'(x) = -4x + 1. \] Now, evaluating the derivative at \( a = 1 \): \[ f'(1) = -4(1) + 1 = -4 + 1 = -3. \] Now, we use the formula for the linear approximation: \[ L(x) = f(a) + f'(a)(x - a). \] Substituting the values, we get: \[ L(x) = -1 - 3(x - 1). \] This simplifies to: \[ L(x) = -1 - 3x + 3 = 2 - 3x. \] Thus, the linear approximation is: \[ L(x) = 2 - 3x. \] Now, here’s a fun fact about linear approximations: they are like the “best friends” of calculus! Just like how your best friend can help you navigate through life’s complexities, the linear approximation helps us simplify complicated functions near a specific point so we can understand their behavior better. If you're itching to dive deeper into the world of calculus, there's a treasure trove of resources out there! Books like "Calculus Made Easy" by Silvanus P. Thompson can make concepts feel less daunting, and online platforms such as Khan Academy offer interactive lessons that make learning calculus engaging and super fun!

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