Front

What is the linear approximation of a function at a point?

Back

The linear approximation of a function at a point is the value of the function at that point plus the product of the derivative at that point and the change in x. It is given by the formula: \( L(x) = f(a) + f'(a)(x - a) \) where \( a \) is the point of approximation.

Front

How do you estimate \( \sqrt{23} \) using linear approximation at \( x = 25 \)?

Back

To estimate \( \sqrt{23} \) using linear approximation at \( x = 25 \), first find \( f(25) = 5 \) and \( f'(x) = \frac{1}{2\sqrt{x}} \) which gives \( f'(25) = \frac{1}{10} \). Then use the formula: \( L(23) = f(25) + f'(25)(23 - 25) = 5 - \frac{1}{10} \cdot 2 = 4.8 \).

Front

What is the formula for the derivative of a function?

Back

The derivative of a function \( f(x) \) at a point \( x \) is defined as: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

Front

How do you find the linear approximation of a differentiable function at a point?

Back

To find the linear approximation of a differentiable function \( f \) at a point \( a \), use the formula: \( L(x) = f(a) + f'(a)(x - a) \).

Front

What is the significance of the tangent line in linear approximation?

Back

The tangent line at a point on the graph of a function provides the best linear approximation of the function near that point.

Front

Given \( f(-2) = -4 \) and \( f'(x) = -2x - 1 \), how do you approximate \( f(-1.95) \)?

Back

Using the linear approximation: \( L(x) = f(-2) + f'(-2)(x + 2) \). Calculate \( f'(-2) = 3 \), then \( L(-1.95) = -4 + 3(0.05) = -3.85 \).

Front

What is the local linear approximation of a function?

Back

The local linear approximation of a function is the approximation of the function's value near a specific point using the tangent line at that point.

Front

How do you use linear approximation to estimate \( f(4.8) \) if \( f(5) = 3 \) and \( f'(5) = 4 \)?

Back

Use the formula: \( L(4.8) = f(5) + f'(5)(4.8 - 5) = 3 + 4(-0.2) = 2.2 \).

Front

What is the derivative of \( f(x) = x^3 - x \)?

Back

The derivative is \( f'(x) = 3x^2 - 1 \).

Front

How do you estimate \( f(2.5) \) for \( f(x) = x^3 - x \) using linear approximation at \( x = 2 \)?

Back

First, find \( f(2) = 2^3 - 2 = 6 \) and \( f'(2) = 3(2^2) - 1 = 11 \). Then use \( L(2.5) = f(2) + f'(2)(2.5 - 2) = 6 + 11(0.5) = 11.5 \).