Question

Find the linearization L of the following functions at the indicated points: (a) f(x) = √(x^2 + 9), x = -4 (b) s(t) = t + 1/t, t = 1.

Answer 1

To find the linearization of a function at a given point, find the tangent line at that point by taking the derivative and using the point-slope formula. For the function f(x) = √(x^2 + 9) and x = -4, the equation of the tangent line is y = (-8/5)x + 28/5.

Step-by-step explanation:

To find the linearization of a function at a given point, we can use the equation for the tangent line at that point. For the function f(x) = √(x^2 + 9) and x = -4, we need to find the equation of the tangent line at x = -4.

To do this, we find the slope of the tangent line by taking the derivative of the function at x = -4. The derivative of f(x) = √(x^2 + 9) is f'(x) = (2x) / √(x^2 + 9). Evaluating this at x = -4, we get f'(-4) = (2(-4)) / √((-4)^2 + 9) = -8 / √25 = -8/5.

Now we can use the point-slope formula to find the equation of the tangent line. The point-slope formula is y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope. Plugging in x1 = -4, y1 = f(-4) = √((-4)^2 + 9) = √25 = 5, and m = -8/5, we get y - 5 = (-8/5)(x - (-4)). Simplifying this equation gives y = (-8/5)x + 28/5. This is the equation of the tangent line at x = -4.