Problem 112 - Calculus and its applications

Question

Then estimate the \(x\) -values at which tangent lines are horizontal. $$ f(x)=x^{4}-3 x^{2}+1 $$

Step-by-Step Solution

Answer

Tangent lines are horizontal at \( x = 0 \), \( x = \sqrt{\frac{3}{2}} \), and \( x = -\sqrt{\frac{3}{2}} \).

Step 1: Understand the Problem

We need to find the values of \( x \) where the tangent lines to the curve \( f(x)=x^{4}-3x^{2}+1 \) are horizontal. The slope of a tangent line is given by the derivative of the function.

Step 2: Find the First Derivative

Calculate the derivative of the function \( f(x) \).\[ f'(x) = \frac{d}{dx}(x^4 - 3x^2 + 1) = 4x^3 - 6x \]

Step 3: Set the Derivative to Zero

To find where the tangent line is horizontal, set the derivative equal to zero and solve for \( x \).\[ 4x^3 - 6x = 0 \]

Step 4: Factor the Derivative

Factor the equation to solve for \( x \).\[ 4x(x^2 - \frac{3}{2}) = 0 \]This gives us two factors: \( 4x = 0 \) and \( x^2 - \frac{3}{2} = 0 \).

Step 5: Solve for Critical Points

Solve each factor for \( x \).- From \( 4x = 0 \), we get \( x = 0 \).- From \( x^2 - \frac{3}{2} = 0 \), we solve \( x^2 = \frac{3}{2} \), giving \( x = \pm \sqrt{\frac{3}{2}} \).

Step 6: State the Solutions

The \( x \)-values at which the tangent lines to the curve \( f(x)=x^{4}-3x^{2}+1 \) are horizontal are \( x = 0 \), \( x = \sqrt{\frac{3}{2}} \), and \( x = -\sqrt{\frac{3}{2}} \).

Key Concepts

Horizontal Tangents

Understanding horizontal tangents is crucial when analyzing the behavior of a function's graph. In calculus, a horizontal tangent line indicates where the slope of the curve is zero. That means, at that specific point, the function doesn't have an upward or downward tilt, it's momentarily flat.
To find where a tangent line is horizontal, we need to find the derivative of the function and then set that derivative equal to zero. This is because the derivative of a function represents the slope of the tangent line. If the derivative is zero, the slope is zero, which denotes a horizontal position.
Horizontal tangents often occur at peaks, valleys, or points of inflection in a graph. These are critical points where the function may be changing direction or behavior. It's essential to correctly identify these spots as they provide significant insights into the nature of the function itself.

Critical Points

Critical points are the x-values on a function where its derivative is zero or undefined. These points are essential because they usually indicate where the graph of a function changes direction, either forming a peak or a valley.
For the function given: \( f(x)=x^4 - 3x^2 + 1 \), the critical points were found by setting the derivative \( f'(x) = 4x^3 - 6x \) to zero, resulting in the equation \( 4x(x^2 - \frac{3}{2}) = 0 \). This equation gives us critical points at \( x = 0 \), \( x = \sqrt{\frac{3}{2}} \), and \( x = -\sqrt{\frac{3}{2}} \).
Critical points are vital in determining the intervals of increase or decrease of a function as well as identifying local maximum or minimum values. After finding the critical points, you should analyze the function around these points to determine their specific nature regarding whether they are maxima or minima.

Derivative Calculation

Calculating derivatives is a fundamental skill in calculus as it allows us to understand how a function changes at any given point. The derivative tells us the rate of change or the slope of the tangent line at a particular point on the curve.
For the function \( f(x) = x^4 - 3x^2 + 1 \), finding the first derivative involves using the power rule: differentiate each term separately. Here, the derivative is \( f'(x) = 4x^3 - 6x \).
Once the derivative is calculated, it serves as a tool to analyze the behavior of the function. By studying the derivative, you can find where the function is increasing, decreasing, or where it has horizontal tangents. Derivative calculation is also the first step towards deeper topics like optimization and using the second derivative to determine concavity.