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Find the critical points and the local extreme values. $$f(x)=\sin ^{2} x-\sqrt{3} \sin x, \quad 0

Short Answer

Expert verified
The critical points of the function are \(x = \frac{\pi}{3}\) and \(x = \frac{\pi}{2}\). The function has a local minimum at \(x = \frac{\pi}{3}\) and a local maximum at \(x = \frac{\pi}{2}\).

Step by step solution

01

Differentiate the function

Calculate the first derivative of the function \(f(x) = \sin^2x - \sqrt{3}\sin x\). Using the chain rule for differentiation, the derivative of \(\sin^2x\) is \(2\sin x \cos x\), and the derivative of \(\sqrt{3}\sin x\) is \(\sqrt{3}\cos x\). The first derivative \(f'(x)\), becomes \(f'(x) = 2\sin x\cos x - \sqrt{3}\cos x\).
02

Calculate the critical points

The critical points are determined by setting the first derivative equal to zero, \(2\sin x\cos x - \sqrt{3}\cos x = 0\). Factor out \(\cos x\), the equation becomes \(\cos x (2\sin x - \sqrt{3}) = 0\). Thus, the critical points occur when \(\cos x = 0\) or when \(2\sin x = \sqrt{3}\). Solving these equations, in the given interval, we get \(x=\frac{\pi}{2}\) and \(x=\frac{\pi}{3}\) respectively.
03

Second Derivative Test

Take the second derivative of the given function \(f'(x)\). This gives \(f''(x) = 2\cos^2x - 2\sin^2x -\sqrt{3}\sin x\). We will test the sign of \(f''(x)\) at each critical point. For \(x=\frac{\pi}{3}\) we get a positive result, which means that \(x=\frac{\pi}{3}\) is a local minimum. For \(x=\frac{\pi}{2}\) we get a negative result, indicating that \(x=\frac{\pi}{2}\) is a local maximum.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, focused on computing the rate at which a function changes. For a function like \(f(x) = \randomFunction \), finding its derivative \(f'(x)\) is crucial to understanding its behavior.

Imagine you're walking and tracking your progress on a map. Differentiation is like looking at your speed at each moment; it shows how fast you are moving towards your destination. The process involves rules and operations that allow you to deduce the steepness (or 'slope') of the function at any given point.

When we differentiate the function \(f(x) = \randomFunction\) and obtain the first derivative, we're effectively calculating a new function that provides the slope of the tangent line at any point on \(f(x)\). This slope often answers questions about increasing or decreasing trends and can predict the function's future behavior.
Chain Rule
The chain rule is a technique in calculus for finding the derivative of a composite function. It's like taking apart a machine to understand how different gears work together. Imagine you have one function inside another; the chain rule is how you untangle this combination to find out how changes in the input affect the output.

In the context of our original problem, the chain rule allowed us to differentiate \(\randomFunction\) smoothly. Applying the chain rule means recognizing the inner function (like \(\randomFunction\) in this case) and the outer function (which can be something like \(\randomFunction\)) and then taking the derivatives step by step to preserve the relationship between them. A helpful analogy is to think of a set of nesting dolls, where each doll must be opened in a specific order to reveal the smallest doll inside.
Second Derivative Test
After finding the critical points of a function, it's natural to wonder if these points are hills (maximums), valleys (minimums), or neither. The second derivative test gives us this insight by checking the curvature of the function at these points.

It works like checking if a path goes up a hill or down into a valley: a positive second derivative suggests a valley (a local minimum), and a negative second derivative implies a hill (a local maximum). A result of zero, however, is inconclusive and may require further investigation. In our exercise, applying the second derivative test at the critical points \(x=\frac{\randomFunction}{\randomFunction}\) and \(x=\frac{\randomFunction}{\randomFunction}\) helped determine which were actually local extremum points.
Local Extreme Values
Local extreme values are the highest or lowest points on a function within a certain range. To continue with previous metaphors, these are the peaks or bottoms of the hills and valleys on your hiking trail. They represent points where the function takes on maximum or minimum values compared to points directly around them.

Using differentiation and the second derivative test is akin to being given a topographical map of the trail. It helps to identify these critical sections so that you can plan accordingly - whether to prepare for a steep climb or to enjoy an easy descent. Finding these local extremes, as seen in our problem where we found the maximum at \(x=\frac{\randomFunction}{\randomFunction}\) and the minimum at \(x=\frac{\randomFunction}{\randomFunction}\), is not just about solving a mathematical puzzle but understanding the landscape of the function we're analyzing.

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Most popular questions from this chapter

(Oblique asymplotes) Let \(r(x)=p(x) / q(x)\) be a rational function. If (degree of \(p)=(\text { degree of } q)+1,\) then \(r\) can be Written in the form \(r(x)=a x+b+\frac{Q(x)}{q(x)}\) with \((\text { degree } Q)<(\text { degree } q)\). Show that \([r(x)-(a x+b)] \rightarrow 0\) both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty .\) Thus the graph of \(f\) "approaches the line \(y=$$a x+b\)" both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty\). The line \(y=\) \(a x+b\) is called an oblique asymptote.

An object moves along a coordinate line, its position at each time \(t \geq 0\) being given by \(x(t)\). Find the times \(t\) at which the object changes direction. $$x(t)=\left(t^{2}-8 t+15\right)^{3}$$.

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. An object projected vertically upward passes every height less than the maximum twice, once on the way up and once on the way down. Show that the speed is the same in each direction. Measure height in feet.

The path of a ball is the curve \(y=m x-\frac{1}{4}\left(m^{2}, 1\right) x^{2}\) Here the origin is taken as tie point from which the ball is thrown and \(m\) is the initial slope of the trajectory. \(\Lambda\) ta distance which depends on \(m\). tie ball returns :o the height from which it was thrown. What value of \(m\) maximizes this distance'?

Water is dripping through the bottom of a conical cup 4 inches across and 6 inches deep. Given that the cup loses half a cubic inch of water per minute, how fast is the water level dropping when the water is 3 inches deep?

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