Question

Multiple Choice Find the instantaneous rate of change of \(f(x)=x^{2}-2 / x+4\) at \(x=-1 .\) $$(\mathbf{A})-7 \quad(\mathbf{B})-4 \quad(\mathbf{C}) 0 \quad(\mathbf{D}) 4$$

Short answer

The instantaneous rate of change of the function \(f(x) = x^{2}-2 / x+4\) at \(x = -1\) is approximately -1.78. None of the given options (A, B, C, D) match this result.

Step-by-step solution

Find the Derivative

The derivative of a function gives us the rate of change of the function. Let's calculate the derivative of the provided function \(f(x) = x^{2}-2 / x+4\). First, we need to properly interpret the given function as \(f(x) = x^{2} - (2 / (x+4))\). We will apply power rule for \(x^{2}\) which states that the derivative of \(x^{n}\) is \(n*x^{n-1}\), and will apply the quotient rule for \(- (2 / (x+4))\) which states that the derivative of \(u/v\) is \((vu'-uv')/v^{2}\), where \(u'\) and \(v'\) are derivatives of \(u\) and \(v\) respectively.

Compute the Derivative

The derivative of the function \(f(x) = x^{2} - (2 / (x+4))\) is calculated as follows: The derivative of \(x^{2}\) according to power rule is \(2x\). The derivative of \(- (2 / (x+4))\) according to quotient rule is: \(v = x+4\), \(u = -2\), \(v' = 1\), \(u' = 0\). Substitute these into quotient rule, we get: \((u'v - uv')/v^{2} = (0 - (1*-2))/((x+4)^{2}) = 2/(x+4)^{2}\). Hence, the derivative \(f'(x) = 2x + 2/(x+4)^{2}\).

Find the Instantaneous Rate of Change

The instantaneous rate of change at \(x = -1\) can be found by substituting \(x = -1\) into \(f'(x)\). Therefore, the instantaneous rate of change at \(x = -1\) is \(f'(-1) = 2*-1 + 2/(((-1)+4)^{2}) = -2 + 2/3^{2} = -2 + 2/9 = -2 + 0.22 approximately = -1.78\). Hence, none of the given options match, implying that there may be a mistake either in the question or the solution.

Key concepts

Derivative of a Function

Understanding the derivative of a function is essential for grasping the concept of instantaneous rate of change. In calculus, the derivative represents the slope of the tangent line to the curve of a function at a given point. This slope essentially tells us how the function is changing at that particular instant; hence it is also called the instantaneous rate of change.

Whenever you calculate the derivative, you're finding a formula that provides the rate of change for any point on the original function. It's like having a speedometer that works for every moment of a journey, not just at one specific time.

Real-world Analogies

Imagine you're driving a car and your speed is constantly changing. The speedometer gives you the rate of change of your distance over time - your instantaneous speed. Similarly, derivatives provide the rate of changing output values of the function with respect to changes in input values.

Power Rule Differentiation

The power rule is a quick and handy tool for finding derivatives of functions that involve powers of the variable, such as polynomials. According to this rule, if you have a function like \(y = x^n\), where \(n\) is any real number, the derivative of that function is \(y' = nx^{n-1}\).

For example, when you have \(f(x) = x^2\), applying the power rule gives you the derivative \(f'(x) = 2x^1\), or simply \(2x\). This rule significantly simplifies the process of differentiation since you can apply it directly without rewriting the function.

Application in Exercises

When faced with complex functions, identify the terms that can be handled with the power rule. This can make the whole process of finding derivatives more approachable and less time-consuming.

Quotient Rule Differentiation

The quotient rule is another fundamental tool in calculus. It's used when differentiating quotients of functions, i.e., one function divided by another. The rule states that if you have two differentiable functions \(u(x)\) and \(v(x)\), the derivative of the quotient \(u(x)/v(x)\) is \((v(x)u'(x) - u(x)v'(x))/v(x)^2\), where \(u'(x)\) and \(v'(x)\) are the derivatives of \(u(x)\) and \(v(x)\) respectively.

Let's simplify that: when you have a fraction to differentiate, you take the bottom function times the derivative of the top function, subtract the top function times the derivative of the bottom function, and then divide all that by the square of the bottom function.

Quotient Rule in Action

If confronted with a function like \(f(x) = \frac{2}{x + 4}\), you'd identify \(u(x) = 2\) and \(v(x) = x + 4\), apply the quotient rule, and simplify. The careful application of this rule is crucial, as mistakes in this step can lead to entirely incorrect solutions.