Let $h$ be a differentiable function with $h(-2)=-7$ and $h^{\prime}(-2)=-8$.
What is the value of the approximation of $h(-1.9)$ using the function's local linear approximation at $x=-2$ ?

Question

Let $h$ be a differentiable function with $h(-2)=-7$ and $h^{\prime}(-2)=-8$.
What is the value of the approximation of $h(-1.9)$ using the function's local linear approximation at $x=-2$ ?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Use the formula for local linear approximation, which is $L(x) = f(a) + f'(a)(x - a)$. ‖ ‖ step 2 ⋮ Given that $h(-2) = -7$ and $h'(-2) = -8$, we can substitute these values into the formula with $a = -2$. ‖ ‖ step 3 ⋮ Calculate the local linear approximation at $x = -1.9$: $L(-1.9) = h(-2) + h'(-2)(-1.9 - (-2))$. ‖ ‖ step 4 ⋮ Simplify the expression: $L(-1.9) = -7 + (-8)(-1.9 - (-2))$. ‖ ‖ step 5 ⋮ Calculate the value: $L(-1.9) = -7 + (-8)(0.1) = -7 - 0.8 = -7.8$. ‖ ∻Answer∻ ⚹ $h(-1.9) \approx -7.8$ ⚹ ∻Key Concept∻ ⚹Local Linear Approximation⚹ ∻Explanation∻ ⚹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.⚹◊What is the formula for the local linear approximation of a function at a point $x=a$?⍭ Generate me a similar question◊

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