/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 a) find the coordinates of any s... [FREE SOLUTION] | 91Ó°ÊÓ

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a) find the coordinates of any stationary points for the graph of the equation b) state, with reasoning, whether each stationary point is a minimum, maximum or neither c) sketch a graph of the equation and indicate the coordinates of each stationary point on the graph. $$y=x-\sqrt{x}$$

Short Answer

Expert verified
The stationary point is \( \left( \frac{1}{4}, -\frac{1}{4} \right) \), and it is a minimum.

Step by step solution

01

Find the first derivative

To find the stationary points, start by calculating the first derivative of the function. Given \( y = x - \sqrt{x} \), rewrite \( \sqrt{x} \) as \( x^{1/2} \). The derivative is \( \frac{dy}{dx} = 1 - \frac{1}{2}x^{-1/2} \), which simplifies to \( 1 - \frac{1}{2\sqrt{x}} \).
02

Set the first derivative to zero

Find the stationary points by setting the first derivative equal to zero. Solve the equation \( 1 - \frac{1}{2\sqrt{x}} = 0 \). Rearranging gives \( \frac{1}{2\sqrt{x}} = 1 \), thus \( 2\sqrt{x} = 1 \). Squaring both sides, we get \( 4x = 1 \), so \( x = \frac{1}{4} \).
03

Calculate the y-coordinate of the stationary point

Using \( x = \frac{1}{4} \) in the original equation \( y = x - \sqrt{x} \), substitute to find the y-coordinate. This gives \( y = \frac{1}{4} - \sqrt{\frac{1}{4}} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \). Therefore, the stationary point is \( \left( \frac{1}{4}, -\frac{1}{4} \right) \).
04

Determine the nature of the stationary point using the second derivative

Find the second derivative to determine if the stationary point is a maximum or minimum. The second derivative is \( \frac{d^2y}{dx^2} = \frac{1}{4}x^{-3/2} = \frac{1}{4\sqrt{x^3}} \). Substitute \( x = \frac{1}{4} \) into the second derivative: \( \frac{1}{4\sqrt{(1/4)^3}} \), which simplifies to \( 2 \). Since this value is positive, the stationary point at \( \left( \frac{1}{4}, -\frac{1}{4} \right) \) is a minimum point.
05

Sketch the Graph

Sketch the graph of \( y = x - \sqrt{x} \). The curve starts from the origin (since both terms are zero at \( x = 0 \)), increasing with the slope determined by the derivative. Mark the minimum point we calculated at \( \left( \frac{1}{4}, -\frac{1}{4} \right) \) on the graph. The graph continues to increase to the right of this point

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stationary Points
Stationary points on a graph are where the slope of the tangent to the curve is zero. This means that at these points, the curve neither rises nor falls for a tiny region around it. To find these points, we must first take the derivative of the function and set it equal to zero. The derivative tells us the slope of the tangent line at any given point. Therefore, by solving the equation resulting from setting the first derivative to zero, we can determine the x-values of stationary points. For example, if we have a function like \( y = x - \sqrt{x} \), finding the stationary point involves differentiating and solving \( \frac{dy}{dx} = 0 \). This helps us find where the turning points occur on the curve.
First Derivative
The first derivative of a function, often denoted as \( \frac{dy}{dx} \), indicates the rate at which the function's value changes with respect to changes in \( x \). It gives us the slope of the tangent line to the function's graph at any point. In simpler terms, the first derivative tells us how steep the graph is and in which direction it is moving. For a function like \( y = x - \sqrt{x} \), the first derivative is \( 1 - \frac{1}{2\sqrt{x}} \). By setting this equal to zero, we can solve for \( x \) to find the stationary points. Any value of \( x \) that makes the first derivative zero indicates a turning point of the function.
Second Derivative
The second derivative of a function provides us with information about the curve's concavity; that is, it helps us understand whether the graph is curving upwards or downwards at a particular point. This is essential for determining the nature of stationary points. If the second derivative, noted as \( \frac{d^2y}{dx^2} \), is positive at a stationary point, the curve is concave up, which indicates a local minimum. Conversely, if the second derivative is negative, the curve is concave down, signifying a local maximum. In our earlier function \( y = x - \sqrt{x} \), using the second derivative \( \frac{1}{4\sqrt{x^3}} \) and substituting \( x = \frac{1}{4} \), we found it to be positive, confirming a minimum point at this stationary point.
Minimum and Maximum Points
Minimum and maximum points, also called extrema, refer to points on a graph where the function reaches its lowest or highest value, respectively, in a neighborhood. Finding these points involves using both the first and second derivatives. Once a stationary point is found using the first derivative, the second derivative test helps determine whether it is a minimum or maximum. A positive second derivative suggests a minimum, while a negative indicates a maximum. In the function \( y = x - \sqrt{x} \), our calculations showed a stationary point at \( \left( \frac{1}{4}, -\frac{1}{4} \right) \), with a positive second derivative, thus indicating a minimum.

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Most popular questions from this chapter

Consider the function \(f(x)=(1+x)^{2}(5-x)\) a) Show that the line tangent to the graph of \(f\) where \(x=1\) does not intersect the graph of the function again. b) Also show that the tangent line at (0,5) intersects the graph of \(f\) at a turning point. c) Sketch the graph of \(f\) and the two tangents from a) and b).

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