Problem 9 - Calculus Early Transcendentals

Question

Locate the critical points and identify which critical points are stationary points. $$ f(x)=\frac{x+1}{x^{2}+3} $$

Step-by-Step Solution

Answer

The critical points are \(x = -3\) and \(x = 1\), both are stationary points.

Step 1: Find the First Derivative

To find the critical points of the function, we need to first find its derivative. The function to differentiate is \(f(x) = \frac{x+1}{x^2+3}\). Using the quotient rule, \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), we identify \(u = x+1\) and \(v = x^2+3\). The derivatives are \(u' = 1\) and \(v' = 2x\). Thus, the derivative is:\[f'(x) = \frac{(1)(x^2+3) - (x+1)(2x)}{(x^2+3)^2} = \frac{x^2 + 3 - 2x^2 - 2x}{(x^2+3)^2} = \frac{-x^2 - 2x + 3}{(x^2+3)^2}\]

Step 2: Set Derivative to Zero

To find the critical points, set \(f'(x) = 0\). This occurs when the numerator is zero:\[-x^2 - 2x + 3 = 0\]This is a quadratic equation that can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a = -1\), \(b = -2\), \(c = 3\).

Step 3: Solve Quadratic Equation

Substitute \(a = -1\), \(b = -2\), and \(c = 3\) into the quadratic formula:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(3)}}{2(-1)}\]\[x = \frac{2 \pm \sqrt{4 + 12}}{-2}\]\[x = \frac{2 \pm \sqrt{16}}{-2}\]\[x = \frac{2 \pm 4}{-2}\]This gives the solutions:\(x = -3\) and \(x = 1\).

Step 4: Identify Stationary Points

Critical points occur where the derivative is zero or undefined. Since we have found the solutions \(x = -3\) and \(x = 1\) by setting the numerator of the derivative to zero, both are stationary points because the denominator \((x^2+3)^2\) is never zero for real \(x\). Therefore, the critical points \(x = -3\) and \(x = 1\) are stationary points.

Key Concepts

Stationary Points

In calculus, stationary points are points on a graph where the derivative of a function is zero. This indicates that the function is not changing at those points—meaning it could be a maximum, minimum, or saddle point where the slope of the tangent is horizontal. Identifying stationary points is crucial for understanding the behavior of a function's graph.

To find these points, you first take the derivative of the function and set it to zero. Solving this equation gives you the x-values of the stationary points. However, a function's derivative can also be zero or undefined. If it is undefined, these points are still critical but not necessarily stationary unless the derivative equals zero.

If we look at an example where a given function is \(f(x) = \frac{x+1}{x^2+3}\), first, identify where \(f'(x) = 0\) to find stationary points. The denominator in this scenario does not equal zero for real \(x\), so the stationary points are indeed given by the solutions to the numerator of the derivative.

Quotient Rule

The quotient rule is an essential differentiation technique used when dealing with functions that are the ratio of two differentiable functions. It helps find the derivative of a function like \(f(x) = \frac{u}{v}\), where \(u\) and \(v\) are both differentiable functions.

The quotient rule formula states: \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\). This means you take the derivative of the numerator \(u'\) times the denominator \(v\), subtract from it the product of the numerator \(u\) and the derivative of the denominator \(v'\), then divide the whole expression by the square of the denominator.

Consider using the quotient rule for \(f(x) = \frac{x+1}{x^2+3}\):

• The top function \(u = x+1\) which has a derivative \(u' = 1\).
• The bottom function \(v = x^2+3\) which has a derivative \(v' = 2x\).

By plugging these into the quotient rule formula, you can determine the derivative \(f'(x)\), which is essential for locating stationary points.

Quadratic Formula

The quadratic formula provides a straightforward method for solving quadratic equations of the form \(ax^2 + bx + c = 0\). This is crucial when finding roots, or zero points, of polynomial equations that appear while setting a derivative to zero for locating critical points.

The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. Solving for \(x\) will yield possible values that may represent points where the function's rate of change is zero.

Let's say you find a derivative like: \(-x^2 - 2x + 3 = 0\). Using \(a = -1\), \(b = -2\), and \(c = 3\) in the quadratic formula gives us:

• Calculate the discriminant: \(b^2 - 4ac = 4 + 12 = 16\).
• Root values: \(x = \frac{2 \pm 4}{-2}\).

Thus, \(x = -3\) or \(x = 1\) are the solutions and in this case, represent the stationary points. This highlights the power of the quadratic formula in bridging derivatives to find meaningful x-values.