Problem 17 - Calculus

Question

For \(y=x^{3}+4 x^{2}-3 x+1,\) find all points where the tangent is horizontal.

Step-by-Step Solution

Answer

The points with horizontal tangents are \((\frac{1}{3}, \frac{13}{27})\) and \((-3, 19)\).

Step 1: Find the derivative

To determine where the tangent is horizontal, we need to find the derivative of the function. The derivative will tell us the slope of the tangent line at any given point. Let's compute the derivative of the function: \[ y = x^3 + 4x^2 - 3x + 1 \] The derivative of each term is: - The derivative of \(x^3\) is \(3x^2\). - The derivative of \(4x^2\) is \(8x\). - The derivative of \(-3x\) is \(-3\). - The derivative of the constant \(1\) is \(0\). Putting it all together, the derivative is: \[ y' = 3x^2 + 8x - 3 \]

Step 2: Set the derivative equal to zero

To find the points where the tangent is horizontal, set the derivative equal to zero and solve for \(x\) because a horizontal tangent corresponds to a slope of zero:\[ 3x^2 + 8x - 3 = 0 \]

Step 3: Solve the quadratic equation

The equation \(3x^2 + 8x - 3 = 0\) is a quadratic equation. We'll use the quadratic formula to solve it, which is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation, \(a = 3\), \(b = 8\), and \(c = -3\). Plug these into the formula:\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]\[ x = \frac{-8 \pm \sqrt{64 + 36}}{6} \]\[ x = \frac{-8 \pm \sqrt{100}}{6} \] \[ x = \frac{-8 \pm 10}{6} \] This gives two solutions:\[ x = \frac{2}{6} = \frac{1}{3} \] \[ x = \frac{-18}{6} = -3 \]

Step 4: Find corresponding y-values

Now that we have the \(x\)-values, we need the corresponding \(y\)-values to find the actual points:1. For \(x = \frac{1}{3}\), substitute into the original function:\[ y = \left(\frac{1}{3}\right)^3 + 4\left(\frac{1}{3}\right)^2 - 3\left(\frac{1}{3}\right) + 1 \] \[ y = \frac{1}{27} + \frac{4}{9} - 1 + 1 \] \[ y = \frac{1 + 12 - 27 + 27}{27} = \frac{13}{27} \] 2. For \(x = -3\), substitute into the original function:\[ y = (-3)^3 + 4(-3)^2 - 3(-3) + 1 \] \[ y = -27 + 36 + 9 + 1 \] \[ y = 19 \] Thus, the points are \(\left(\frac{1}{3}, \frac{13}{27}\right)\) and \((-3, 19)\).

Key Concepts

Derivative

In calculus, the derivative of a function at a given point determines the slope of the tangent line to the function's graph at that point. It tells us how fast the function's value is changing with respect to changes in the input. To find the derivative, we use rules for differentiation on each term of the function. For instance, if we have a polynomial function like \[ y = x^3 + 4x^2 - 3x + 1 \]we can differentiate each term:

• The term \(x^3\) becomes \(3x^2\).
• The term \(4x^2\) becomes \(8x\).
• The term \(-3x\) becomes \(-3\).
• Constant terms like \(1\) become 0.

After applying these differentiation rules, we obtain the derivative:\[ y' = 3x^2 + 8x - 3 \]This function \(y'\) can now be used to find important properties about the original function, such as identifying points where the tangent is horizontal.

Quadratic Equation

A quadratic equation is a mathematical expression of the form \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants. Quadratic equations can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows us to find the solutions for \(x\), called the roots or zeroes of the quadratic equation. Solving these equations is crucial when finding points where a derivative equals zero, as these solutions indicate where the slope is zero, resulting in horizontal tangents. For our particular problem \[ 3x^2 + 8x - 3 = 0 \]we plug in \(a = 3\), \(b = 8\), and \(c = -3\) to find that the solutions are:\[ x = \frac{1}{3} \] and \[ x = -3 \]These solutions help in pinpointing the exact locations along the curve where horizontal tangents are found.

Horizontal Tangent Line

Horizontal tangent lines occur when the slope of the tangent at a point on the curve is zero. This is found by setting the derivative to zero and solving for the \(x\)-values. The slope being zero means that the direction of the slope is perfectly flat – or horizontal – at these points. In terms of the function we are working with, \[ y = x^3 + 4x^2 - 3x + 1 \]the derivative was found to be \[ y' = 3x^2 + 8x - 3 \]By setting this to zero, i.e.,\[ 3x^2 + 8x - 3 = 0 \]we are able to solve for the \(x\) values at which this condition occurs. The solutions \( x = \frac{1}{3} \) and \( x = -3 \) represent the points on the curve where these horizontal tangents exist. Calculating the corresponding \(y\)-values completes the identification of these special points.

Slope of Tangent

The slope of a tangent line to a curve at a given point is the value of the derivative at that point. It describes the steepness and direction of the tangent line. When dealing with a function, differentiating the function allows us to pinpoint the slope at any value of \(x\). For the function: \[ y = x^3 + 4x^2 - 3x + 1 \]we find the derivative \[ y' = 3x^2 + 8x - 3 \]This derivative function tells us how the slope changes at any point.

• If \(y' > 0\), the function is increasing at that point.
• If \(y' < 0\), the function is decreasing at that point.
• If \(y' = 0\), the function is neither increasing nor decreasing, which indicates a horizontal tangent.

Finding where \(y' = 0\) enables identification of positions where the curve levels off, such as at the solutions we calculated earlier: \( x = \frac{1}{3} \) and \( x = -3 \).