Question

let f be the function given by f(x) = 3x². what is the approximation for f(-1.2) found by using the line tangent to the graph of f at x = -1?

answer attempt 1 out of 2

tangent line: y =

approximation: f(-1.2) ≈

when , the second derivative , meaning the graph of f is near the point of tangency, making the value ??? an .

Explanation:

Step1: Find \( f(-1) \) and \( f'(-1) \)

First, the function is \( f(x) = 3x^3 \). The derivative \( f'(x) = 9x^2 \).
Calculate \( f(-1) \): \( f(-1)=3\times(-1)^3 = - 3 \).
Calculate \( f'(-1) \): \( f'(-1)=9\times(-1)^2 = 9 \).

Step2: Find the tangent line equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(-1,-3) \) and \( m = f'(-1)=9 \).
Substitute into the formula: \( y - (-3)=9(x - (-1)) \), which simplifies to \( y + 3=9(x + 1) \), and then \( y=9x+9 - 3=9x + 6 \).

Step3: Approximate \( f(-1.2) \)

To approximate \( f(-1.2) \), we use the tangent line equation \( y = 9x+6 \) and substitute \( x=-1.2 \) into it.
\( y=9\times(-1.2)+6=-10.8 + 6=-4.8 \). So \( f(-1.2)\approx - 4.8 \).

Answer:

Tangent Line: \( y = 9x + 6 \)
Approximation: \( f(-1.2)\approx - 4.8 \)