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Magazine Chapter 15: Problem 8
A curve has equation \(y=x(x-4)^{2}\) a) For this curve, find (i) the \(x\) -intercepts (ii) the coordinates of the maximum point (iii) the coordinates of the point of inflexion. b) Use your answers to part a) to sketch a graph of the curve for \(0 \leqslant x \leqslant 4\) clearly indicating the features you have found in part a).
Short Answer
Expert verified
x-intercepts at (0,0) and (4,0), point of inflection at (8/3, 128/27).
Step by step solution
01
Find x-Intercepts
To find the x-intercepts, set \( y = 0 \) and solve for \( x \) in the equation \( y = x(x-4)^2 \). Factoring the equation gives \( x(x-4)^2 = 0 \). This means \( x = 0 \) and \( (x-4)^2 = 0 \), leading to \( x = 4 \). Therefore, the x-intercepts are at \( x = 0 \) and \( x = 4 \).
02
Determine the Derivative
To find the maximum point, first determine the first derivative \( \frac{dy}{dx} \) using the product rule. \( y = x(x-4)^2 = x(x^2 - 8x + 16) \). Differentiate: \( \frac{dy}{dx} = \frac{d}{dx}[x(x^2 - 8x + 16)] = (1)(x^2 - 8x + 16) + x(2x - 8) = x^2 - 8x + 16 + 2x^2 - 8x \). Simplifying the derivative, \( \frac{dy}{dx} = 3x^2 - 16x + 16 \).
03
Find Critical Points
Set the first derivative equal to zero: \( 3x^2 - 16x + 16 = 0 \). Solve this quadratic equation using the quadratic formula, where \( a = 3 \), \( b = -16 \), \( c = 16 \). The solutions are:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{256 - 192}}{6} = \frac{16 \pm \sqrt{64}}{6} = \frac{16 \pm 8}{6}\]Which gives roots \( x = 4 \) and \( x = \frac{8}{3} \). These are the critical points.
04
Evaluate Second Derivative for Concavity
Compute the second derivative \( \frac{d^2y}{dx^2} \). Differentiate \( 3x^2 - 16x + 16 \) to get \( \frac{d^2y}{dx^2} = 6x - 16 \). Evaluate at \( x = 4 \) and \( x = \frac{8}{3} \).For \( x = 4 \): \( 6(4) - 16 = 24 - 16 = 8 \). Since this is positive, \( x = 4 \) is a minimum point.For \( x = \frac{8}{3} \): \( 6(\frac{8}{3}) - 16 = 16 - 16 = 0 \). This indicates a possible point of inflection.
05
Determine Coordinates of Maximum and Inflexion
The maximum point is calculated where the second derivative changes sign. From Step 3, \( x = 4 \) was a minimum, thus \( x = \frac{8}{3} \) is a point of inflection, not a maximum.Calculate \( y \) at these critical points. At \( x = 0 \), \( y = 0(0-4)^2 = 0 \).At \( x = \frac{8}{3} \), substitute in \( y = x(x-4)^2 \):\[ y = \frac{8}{3}\left(\frac{8}{3} - 4\right)^2 = \frac{8}{3}\left(-\frac{4}{3}\right)^2 = \frac{8}{3} \times \frac{16}{9} = \frac{128}{27} \]So, the point of inflection is \( \left(\frac{8}{3}, \frac{128}{27}\right) \).
06
Sketch the Graph
Use the results from the previous steps to sketch the graph. The x-intercepts are at \( (0, 0) \) and \( (4, 0) \). The minimum point is at \( (4, 0) \) and the point of inflection is at \( \left(\frac{8}{3}, \frac{128}{27}\right) \), which will have an upward curve at the interval considered. Make sure to clearly mark these points and show the curve passing through these points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding x-intercepts
X-intercepts are points where the curve crosses the x-axis on a graph. At these points, the value of the function is zero. In other words, x-intercepts are found by solving the equation \( y = 0 \) in the given function equation. For the curve given by \( y = x(x-4)^2 \), setting \( y = 0 \) leads to the equation \( x(x-4)^2 = 0 \). This equation can be solved by factoring it into \( x = 0 \) and \( (x-4)^2 = 0 \). This gives us two x-intercepts at \( x = 0 \) and \( x = 4 \).
- X-intercepts are crucial for sketching graphs as they indicate where the curve meets the x-axis.
- The order of the factors can affect the curve's nature at these intercepts, like whether the curve bounces off or passes through the axis.
Discovering critical points
Critical points on a graph are where the function鈥檚 derivative equals zero or is undefined. They are important because they help identify maximum, minimum, or inflection points of the function. For the function \( y = x(x-4)^2 \), we first find the derivative \( \frac{dy}{dx} \) using differentiation rules. Once we calculate \( \frac{dy}{dx} = 3x^2 - 16x + 16 \), we set it to zero to find the critical points:\[ 3x^2 - 16x + 16 = 0 \]Use the quadratic formula to solve this, resulting in roots \( x = 4 \) and \( x = \frac{8}{3} \).
- Critical points indicate potential peaks or valleys of the curve.
- Knowing these points helps determine specific graph features like maxima, minima, or inflection points.
Second derivative test for concavity
The second derivative test is a way to determine the concavity of a function at its critical points, which tells us if those points are maxima, minima, or inflection points. We find the second derivative \( \frac{d^2y}{dx^2} \) by differentiating the first derivative. For \( 3x^2 - 16x + 16 \), the second derivative is \( 6x - 16 \).Evaluate this at the critical points:- At \( x = 4 \): \( \frac{d^2y}{dx^2} = 8 \), which is positive, indicating a minimum point.- At \( x = \frac{8}{3} \): \( \frac{d^2y}{dx^2} = 0 \), which suggests a possible point of inflection.
- Positive second derivative at a point indicates the curve is concave up, suggesting a local minimum.
- Negative second derivative indicates the curve is concave down, suggesting a local maximum.
- Zero second derivative may indicate an inflection point where the curve changes concavity.
Exploring the product rule
The product rule is a fundamental tool in calculus used for differentiating products of two functions. For a function \( y = u(x)v(x) \), the product rule states that the derivative \( \frac{dy}{dx} \) is given by:\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \]For the curve \( y = x(x-4)^2 \), treat \( u(x) = x \) and \( v(x) = (x-4)^2 \). By applying the product rule, determine:- \( u'(x) = 1 \)- \( v'(x) = 2(x-4)(1) = 2x - 8 \)Substituting into the product rule gives: \( \frac{dy}{dx} = 1(x^2 - 8x + 16) + x(2x - 8) \)Simplifying, we find \( \frac{dy}{dx} = 3x^2 - 16x + 16 \).
- The product rule is essential for differentiation when dealing with products of functions.
- Used to simplify finding first derivatives, which help locate critical points and analyze the curve's behavior.
