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For the function e–x, the linear approximation around x = 2 is
- a)(3 – x) e–2
- b)1 – x
- c)[3 + 2 2− (1 + 2 ) x] e−2
- d)e –2
Correct answer is 'A'. Can you explain this answer?
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For the function e–x, the linear approximation around x = 2 isa)...
(neglecting higher power of x)
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For the function e–x, the linear approximation around x = 2 isa)...
Linear Approximation of ex around x = 2
The linear approximation of a function f(x) around a point x = a is given by the equation:
f(x) ≈ f(a) + f'(a)(x - a)
where f'(a) is the derivative of f(x) evaluated at x = a.
In this question, we are given the function f(x) = ex and the point x = 2. We need to find the linear approximation of f(x) around x = 2.
Step 1: Find f(2) and f'(2)
f(2) = e2 ≈ 7.389
f'(x) = ex
Therefore, f'(2) = e2 ≈ 7.389
Step 2: Use the linear approximation formula
f(x) ≈ f(2) + f'(2)(x - 2)
Substituting the values of f(2) and f'(2), we get:
f(x) ≈ 7.389 + 7.389(x - 2)
Simplifying the equation, we get:
f(x) ≈ 7.389x - 7.389
Therefore, the linear approximation of ex around x = 2 is:
(a) (3 - x) e2
Note: Option (c) is not correct as it includes the term (1 - 2) which is equal to -1.
The linear approximation of a function f(x) around a point x = a is given by the equation:
f(x) ≈ f(a) + f'(a)(x - a)
where f'(a) is the derivative of f(x) evaluated at x = a.
In this question, we are given the function f(x) = ex and the point x = 2. We need to find the linear approximation of f(x) around x = 2.
Step 1: Find f(2) and f'(2)
f(2) = e2 ≈ 7.389
f'(x) = ex
Therefore, f'(2) = e2 ≈ 7.389
Step 2: Use the linear approximation formula
f(x) ≈ f(2) + f'(2)(x - 2)
Substituting the values of f(2) and f'(2), we get:
f(x) ≈ 7.389 + 7.389(x - 2)
Simplifying the equation, we get:
f(x) ≈ 7.389x - 7.389
Therefore, the linear approximation of ex around x = 2 is:
(a) (3 - x) e2
Note: Option (c) is not correct as it includes the term (1 - 2) which is equal to -1.
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