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The equation of the normal line of the function f(x) = sqrt(x + 2) at x = –1 is

a) y = -2x - 1
b) y = -2x + 1
c) y = 0.5x - 1.5
d) y = 0.5x + 1.5

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Final answer:

The equation of the normal line of the function f(x) = sqrt(x + 2) at x = –1 is y = -2x - 1. To find the equation of the normal line, first, find the derivative of the function.

Step-by-step explanation:

The equation of the normal line of the function f(x) = sqrt(x + 2) at x = –1 is y = -2x - 1 (option a).

To find the equation of the normal line, we first need to find the derivative of the function. The derivative of f(x) = sqrt(x + 2) is f'(x) = 1 / (2 * sqrt(x + 2)).

At x = –1, the slope of the tangent line is f'(-1) = 1 / (2 * sqrt(-1 + 2)) = 1 / (2 * sqrt(1)) = 1 / 2.

Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the tangent line slope. So the slope of the normal line is -2.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can plug in the values x1 = -1, y1 = f(-1) = sqrt(-1 + 2) = 1, and m = -2 to get the equation of the normal line as y - 1 = -2(x + 1).

Simplifying further, y = -2x - 2 + 1, y = -2x - 1.

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