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Posted 3 months ago
Let hh be a differentiable function with h(2)=7h(-2)=-7 and h(2)=8h^{\prime}(-2)=-8.
What is the value of the approximation of h(1.9)h(-1.9) using the function's local linear approximation at x=2x=-2 ?
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Use the formula for local linear approximation, which is L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x - a)
step 2
Given that h(2)=7h(-2) = -7 and h(2)=8h'(-2) = -8, we can substitute these values into the formula with a=2a = -2
step 3
Calculate the local linear approximation at x=1.9x = -1.9: L(1.9)=h(2)+h(2)(1.9(2))L(-1.9) = h(-2) + h'(-2)(-1.9 - (-2))
step 4
Simplify the expression: L(1.9)=7+(8)(1.9(2))L(-1.9) = -7 + (-8)(-1.9 - (-2))
step 5
Calculate the value: L(1.9)=7+(8)(0.1)=70.8=7.8L(-1.9) = -7 + (-8)(0.1) = -7 - 0.8 = -7.8
h(1.9)7.8h(-1.9) \approx -7.8
Key Concept
Local Linear Approximation
The local linear approximation of a differentiable function at a point is given by the value of the function at that point plus the derivative at that point times the change in x.

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