In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

g(x) = 8 / x², (2, 2)

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To find the slope of the function's graph at the given point, we need to compute the derivative of the function g(x) = 8 / x². This can be rewritten as g(x) = 8x⁻².

Apply the power rule for differentiation: if f(x) = xⁿ, then f'(x) = n * xⁿ⁻¹. For g(x) = 8x⁻², the derivative g'(x) is calculated as g'(x) = -2 * 8 * x⁻³ = -16 / x³.

Evaluate the derivative at the given point x = 2 to find the slope of the tangent line. Substitute x = 2 into g'(x) to get g'(2) = -16 / (2)³.

Now that we have the slope of the tangent line, use the point-slope form of a line to find the equation of the tangent line. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point (2, 2).

Substitute the slope from step 3 and the point (2, 2) into the point-slope form to get the equation of the tangent line. Simplify the equation to express it in the form y = mx + b.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function at a point provides the slope of the tangent line to the graph of the function at that point. It is a fundamental concept in calculus used to determine how a function changes as its input changes. For the function g(x) = 8/x², finding the derivative will help us calculate the slope at the given point (2, 2).

Power Rule

The power rule is a basic derivative rule used to find the derivative of functions in the form of x^n. It states that the derivative of x^n is n*x^(n-1). For the function g(x) = 8/x², we can rewrite it as 8*x^(-2) and apply the power rule to find its derivative, which is essential for determining the slope at the specified point.

Equation of a Tangent Line

The equation of a tangent line to a function at a given point is found using the point-slope form: y - y₁ = m(x - x₁), where m is the slope from the derivative and (x₁, y₁) is the given point. This equation represents the line that just touches the graph of the function at the point without crossing it, providing a linear approximation of the function near that point.

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.g(x) = 8 / x², (2, 2)