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In Exercises \(19-30\) , find the extreme values of the function and where they occur. $$y=\sqrt{3+2 x-x^{2}}$$

Short Answer

Expert verified
The extreme value is \(2\) and it occurs when \(x = 1\).

Step by step solution

01

Finding the Derivative

The first step is to find the derivative of the given function. To do this, use the chain rule and the power rule of derivatives. Differentiating \(y = \sqrt{3 + 2x - x^{2}}\) yields \(y' = \frac{1}{2\sqrt{3 + 2x - x^{2}}} (2 - 2x)\).
02

Solving for Critical Numbers

Equating the derivative to zero, we have \(\frac{1}{2\sqrt{3 + 2x - x^{2}}} (2 - 2x) = 0\). Solving this equation for \(x\) gives us the critical numbers, which are the potential points for local maxima and minima. We arrive at \(x = 1\).
03

Testing the Endpoints and Critical Numbers

Check the extreme values at the endpoints of the function and at the critical numbers. We need to substitute these values into the original function to get the extreme values. Substitute \(x = 1\) in the function to get \(y = \sqrt{3 + 2(1) - (1)^2} = \sqrt{4} = 2\). This is the only point inside the range of \(x\) where the function can exist, so it is both the maximum and the minimum, which is an extreme point.

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Key Concepts

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

Critical Numbers
In the study of calculus, critical numbers play a vital role in determining the extreme values of functions. These numbers are found where the derivative of a function equals zero or does not exist. To find the critical numbers for the function y = \( \sqrt{3 + 2x - x^2} \), the derivative y' is calculated and set to zero. When y' equals zero, it potentially indicates a point of maximum or minimum value—also known as a local extremum.

By solving y' = 0, you find the critical numbers as seen in the exercise, which here specifically gives us x = 1. It's essential to test this critical number along with any endpoints of the domain to determine whether it corresponds to a maximum, minimum, or neither. Finding and analyzing critical numbers help us understand the behavior of a function and where it levels off or peaks, which is fundamental in various practical applications such as optimization problems.
Derivatives in Calculus
The concept of derivatives is foundational in calculus. Derivatives measure the rate at which a function is changing at any given point and are graphically represented by the slope of the tangent line at that point on the function. To find extreme values of a function, the derivative is a powerful tool.

In the given problem, the derivative y' of the function y = \( \sqrt{3 + 2x - x^2} \) is used to determine where the function stops increasing or decreasing and starts to change direction. This points us to the possible locations of maximums and minimums. Understanding how to take derivatives, interpret them, and apply them to problems is crucial for students mastering calculus.
Chain Rule
The chain rule is a rule in calculus for differentiating compositions of functions. It is instrumental when tackling complex functions that are nested within each other, like a function within a function. To apply the chain rule, identify the 'outer' and 'inner' functions in the composition and then apply the derivative accordingly, following the mantra: 'the derivative of the outside times the derivative of the inside.'

In our exercise, the chain rule is applied to differentiate y = \( \sqrt{3 + 2x - x^2} \). The square root represents the 'outer' function, and the quadratic expression 3 + 2x - x2 is the 'inner' function. Using the chain rule, we find y', which is essential for locating the critical numbers and, hence, the function's extreme values. Understanding the chain rule is vital for students as it appears often in calculus problems involving derivatives.
Power Rule
Another fundamental principle of calculus is the power rule for derivatives. This rule simplifies the process of finding the derivative of a function that is a power of x. It states that if you have a function y = xn, then the derivative y' is nxn-1.

In the given problem, to find the derivative of the square root (which is the same as raising to the power of 1/2), the power rule is invoked after applying the chain rule. The derivative inside the square root, (2 - 2x), also follows the power rule since each term is a power of x. Understanding the power rule allows you to quickly and effectively differentiate a vast array of functions, which is especially useful when finding extreme values of functions.

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Most popular questions from this chapter

Multiple Choice A cylindrical rubber cord is stretched at a constant rate of 2 \(\mathrm{cm}\) per second. Assuming its volume does no change, how fast is its radius shrinking when its length is 100 \(\mathrm{c}\) and its radius is 1 \(\mathrm{cm} ?\) $$\begin{array}{ll}{\text { (A) } 0 \mathrm{cm} / \mathrm{sec}} & {\text { (B) } 0.01 \mathrm{cm} / \mathrm{sec}} 67 {\text{ (C) } 0.02 \mathrm{cm} / \mathrm{sec}}$\\\ {\text { (D) } 2 \mathrm{cm} / \mathrm{sec}} & {\text { (E) } 3.979 \mathrm{cm} / \mathrm{sec}}\end{array}

Quartic Polynomial Functions Let \(f(x)=\) \(a x^{4}+b x^{3}+c x^{2}+d x+e\) with \(a \neq 0\) (a) Show that the graph of \(f\) has 0 or 2 points of inflection. (b) Write a condition that must be satisfied by the coefficients if the graph of \(f\) has 0 or 2 points of inflection.

Multiple Choice If \(y=\tan x, x=\pi,\) and \(d x=0.5,\) what does \(d y\) equal? \(\begin{array}{lll}{\text { (A) }-0.25} & {\text { (B) }-0.5} & {\text { (C) } 0} & {\text { (D) } 0.5}\end{array}\) (E) 0.25

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .

Expanding Circle The radius of a circle is increased from 2.00 to 2.02 \(\mathrm{m} .\) (a) Estimate the resulting change in area. (b) Estimate as a percentage of the circle's original area.

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