Question

Math

Posted 6 months ago

```
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 $x=-2$ ?
```

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Answer from Sia

Posted 6 months ago

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.

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