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Sketch the graph of the function using the approach presented in this section. $$f(x)=2 \sqrt{x}-x, \quad x \in[0,4]$$

Short Answer

Expert verified
The function \(f(x) = 2\sqrt{x} - x\) decreases from \(x = 0\) to \(x = 1\), then increases from \(x = 1\) to \(x = 4\). The range of \(f(x)\) over the given domain is found by evaluating the function at each end point and at the critical point \(x = 1\). The minimum value is 0 at \(x = 4\) and the maximum value is 2 at \(x = 0\).

Step by step solution

01

Identify the Domain and Range

The domain of \(f(x) = 2\sqrt{x} - x\) is \(x \in [0,4]\) as given by the exercise. The range can be found by finding the maximum and minimum values of the function in the given domain, which will be done in the following steps.
02

Find the Critical Points

The critical points occur where the derivative is zero or undefined. First find the derivative of the function: \(f'(x) = \frac{2}{2\sqrt{x}} - 1 = \frac{1}{\sqrt{x}} - 1\). We set \(f'(x) = 0\) to find the critical point(s): \(\frac{1}{\sqrt{x}} - 1 = 0\), thus \(\sqrt{x} = 1\) which implies \(x = 1\). Since this value is in the domain, \(x = 1\) is a critical point.
03

Determine Where the Function Increases and Decreases

Using the critical point and endpoints, divide the domain into intervals: \([0, 1)\), \((1, 4]\). If \(f'(x) > 0\) on an interval, then \(f(x)\) is increasing there. If \(f'(x) < 0\) on an interval, then \(f(x)\) is decreasing. Evaluating the derivative within each interval (e.g. at \(x = 0.5\), \(x = 2\)) confirms that \(f(x)\) is decreasing on \([0, 1)\) and increasing on \((1, 4]\).
04

Plot the Function

On a graph, mark the critical point and endpoints, then sketch the function. It decreases from \(x = 0\) down to \(x = 1\), then increases from \(x = 1\) up to \(x = 4\). Don't forget, the y-values at the critical point and endpoints can be found by substituting these x-values back into the original function; for example, \(f(1) = 2\sqrt{1} - 1 = 1\).

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Key Concepts

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

Critical Points
Critical points of a function are where its derivative is zero or doesn't exist, serving as potential locations where the function's behavior changes. For instance, if you have the function \( f(x) = 2\sqrt{x} - x \), finding its derivative provides us with \( f'(x) = \frac{1}{\sqrt{x}} - 1 \). Setting this derivative to zero gives us \( \sqrt{x} = 1 \), which simplifies to \( x = 1 \). This point is essential because it's where the function could reach a maximum or minimum, or where the shape of the graph might change, like transitioning from increasing to decreasing. It's a bit like a hiking trail: at a critical point, you could be at the peak of a hill (maximum), the bottom of a valley (minimum), or a flat spot where the incline changes.

To ensure this point is genuinely 'critical,' it must lie within the function's domain. Here, the critical point at \( x = 1 \) is within the given domain \( [0,4] \). Thus, our trail guide correctly marks this as a place to watch out for a change in elevation—or in mathematical terms, a possible extreme value on our graph.
Function Domain and Range
The domain of a function defines all the possible input values (x-values), while the range is the collection of all possible output values (y-values) that result from those inputs. It's like setting boundaries for a game: the domain is where you're allowed to play, and the range is the possible scores you can achieve. For the function \( f(x) = 2\sqrt{x} - x \) within the domain \( x \in [0,4] \), we know that these are the x-values we can plug into our equation.

As for the range, it's determined by seeing how high and low the function goes within the domain's limits. This typically involves calculating the function's values at critical points and boundaries. For our function, this means we'd look at \( f(0) \), \( f(1) \), and \( f(4) \) to help identify the range, since the function might bow down or peak at these points, much like a landscape within the confines of our domain.
Derivative of a Function
The derivative of a function is like a speedometer for your car on the graph's landscape—it tells you how fast the function's values are changing at any given point. If the function is \( f(x) = 2\sqrt{x} - x \), then the derivative \( f'(x) \) looks like \( \frac{1}{\sqrt{x}} - 1 \).

When you calculate the derivative, you're essentially finding the slope of the tangent line to the function at any point. If the slope is positive, you're going uphill. If it's negative, you're descending. And if the slope is zero, you've hit a flat spot—a potential peak or valley. Calculating the derivative is like taking a snapshot of your elevation gain or loss at that exact moment. By doing this across the entire domain, you can chart the entire trail map of your function.
Increasing and Decreasing Intervals
Identifying where a function is increasing or decreasing is similar to mapping out which parts of a trail go uphill and which go downhill. For our function \( f(x) = 2\sqrt{x} - x \), we use the derived formula \( f'(x) = \frac{1}{\sqrt{x}} - 1 \) to ascertain where the slope of the graph is positive (increasing) or negative (decreasing).

We divide our domain into intervals, using critical points and endpoints to guide us. For instance, with the intervals \( [0, 1) \) and \( (1, 4] \), we test points in each to see if they make our derivative positive or negative. If, while hiking, you notice you're walking downward towards the critical point, like going from \( x = 0 \) to \( x = 1 \), our function is decreasing. But after the critical point, as you move from \( x = 1 \) to \( x = 4 \), you're heading uphill—our function is increasing. This process helps sketch a more accurate graph of the function, showing us the rises and falls along the way.

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Most popular questions from this chapter

An object moves along the \(x\) -axis, its position at each time \(t\) given by \(x(t)\). Determine those times from \(t=0\) to \(f=2 \pi\) at which the object is moving to the right with increasing speed. $$x(t)=\cos 2 t$$.

The lines \(y=(b / a) x\) and \(y=-(b / a) x\) are called asymtotes of the hyperbola $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$ (a) Draw a figure that illustrates this asymptotic bchavior. (b) Show that the first-quadrant are of the hyperbola, the curve $$y=\frac{b}{a} \sqrt{x^{2}-a^{2}}$$ is indeed asymptotic to the line \(y=(b / a) x\) by showing that $$ \frac{b}{a} \sqrt{x^{2}-a^{2}}-\frac{b}{a} x \rightarrow 0 \text { as } x \rightarrow \infty $$ (c) Procceding as in part (b), show that the second-quadrant are of the hyperbola is asymplotic to the line \(y=\) \(-(b / a) x\) by taking a suitable limit as \(x \rightarrow-\infty\). (The asymptotic behavior in the other quadrants can be verified in an analogous manner, or by appealing to symmetry.)

A horizontal trough 12 feet long has a vertical cross section in the form of a trapezoid. The bottom is 3 feet wide, and the sides are inclined to the vertical at an angle with sine \(\frac{4}{5}\) Given that water is poured into the trough at the rate of 10 cubic feet per minute, how fast is the water level rising when the water is exactly 2 feet deep?

Neglect air resistance. For the numerical calculations, take \(g\) as 32 feet per second \(\mathrm{p}\) er second or as 9.8 meters per second per second. A rubber ball is thrown straight down from a height of 4 feet. If the ball rebounds with one-half of its impact speed and returns exactly to its original height before falling again, how fast was it thrown originally?

Sketch the graph of the function using the approach presented in this section. $$f(x)=\frac{\sin x}{1-\sin x}, \quad x \in(-\pi, \pi)$$

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