/*! 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 61 Tangent lines The graph of \(y... [FREE SOLUTION] | 91Ó°ÊÓ

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Tangent lines The graph of \(y=\left(x^{2}\right)^{x}\) has two horizontal tangent lines. Find equations for both of them.

Short Answer

Expert verified
Answer: There are two horizontal tangent lines, but they coincide, having the same equation: \(y = 1\).

Step by step solution

01

Rewrite the function for simplification

First, let's rewrite the function by simplification: \(y = (x^2)^x = x^{2x}\) Now, we have the function \(y = x^{2x}\).
02

Take the derivative of the function and set it to zero

To find the points where the tangent lines are horizontal, we'll take the derivative of the function with respect to x and set it to zero. We will use the chain rule and exponent rule here: \(y = x^{2x}\) \(\frac{dy}{dx} = \frac{d(x^{2x})}{dx}\) First, let's use the chain rule: \(\frac{dy}{dx} = 2x \cdot x^{2x-1} \cdot \frac{d(2x)}{dx}\) Now, let's find the derivative of \(2x\): \(\frac{d(2x)}{dx} = 2\) So, the derivative of our function is: \(\frac{dy}{dx} = 2x \cdot x^{2x-1} \cdot 2\) Now, set the derivative equal to zero: \(2x \cdot x^{2x-1} \cdot 2 = 0\)
03

Solve the equation for x

Now, we will simplify and solve the equation: \(4x \cdot x^{2x-1} = 0\) We can factor out an \(x\): \(x(4x^{2x-1})=0\) From this equation, we can see that there are two possible values for x: \(x = 0\) and \(x = 1\), since \(x^{2x-1}\) will never be zero for positive values of x.
04

Find corresponding values of y

Now, we will find the corresponding values of y for x=0 and x=1: For \(x = 0\): \(y = (0^2)^0 = 1\) For \(x = 1\): \(y = (1^2)^1 = 1\)
05

Write the equations for the horizontal tangent lines

The horizontal tangent lines will have the same y values since they are both tangent at the same height. We found that the y-value is 1. Therefore, the equations of the horizontal tangent lines are: Horizontal Tangent Line 1: \(y = 1\) Horizontal Tangent Line 2: \(y = 1\) There are two horizontal tangent lines, but they coincide, since they both have the same equation, which is \(y=1\).

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Key Concepts

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

Derivative of Exponential Functions
When working with exponential functions in calculus, understanding how to find their derivatives is essential. Evidently, the derivative represents the rate at which the function's value changes. Exponential functions, which can be generally expressed as \(a^{x}\), where \(a\) is a constant and \(x\) is the exponent, have unique properties.

For functions with a variable as the base and the exponent like \(x^{2x}\), deriving them involves using the concept of logarithmic differentiation. Initially, you would take the natural logarithm of both sides, apply the properties of logarithms to simplify, and then differentiate implicitly. This method is powerful because it transforms a potentially complicated exponential derivative into a more manageable form. Understanding this process is fundamental because exponential functions are inherent in various fields including biology, economics, and physics.
Chain Rule Calculus
The chain rule is one of the most powerful tools in calculus for finding the derivative of composite functions. A composite function is a function composed of two or more functions, where the output of one function becomes the input of another. The rule states that if \(u=g(x)\) and \(y=f(u)\), then the derivative of \(y\) with respect to \(x\) is \(dy/dx = (dy/du) * (du/dx)\).

The chain rule allows calculation of the derivative of complex functions by breaking them down into simpler parts for which derivatives can be more easily calculated. This step-by-step approach makes it easier to solve derivative problems that might first appear intimidating. Especially in the case of finding horizontal tangent lines, where you're often dealing with compositions of functions, the chain rule can simplify the process of finding where the derivative is zero.
Finding Tangent Lines
In calculus, finding the equation of a tangent line to a curve at a given point involves calculating the derivative of the function at that point. The derivative gives the slope of the tangent line, and knowing a point on the line allows us to find its equation using the point-slope form. Specifically, horizontal tangent lines occur where the derivative of the function is zero, signifying no change in the y-direction at those points. These are often critical points that may correspond to local minima or maxima.

In the given exercise, after calculating the derivative and setting it to zero, the resultant x-values are where the function has a horizontal tangent. You then use these x-values to find the corresponding y-values on the original function. Here, this approach led to the conclusion that the curve \(y = x^{2x}\) has a horizontal tangent not at two different heights, as might be typical, but rather coincidentally at the same y-value, yielding identical equations for both horizontal tangent lines: \(y=1\). This revelation highlights the importance of thoroughly analyzing the results when finding tangent lines.

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Most popular questions from this chapter

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(y^{3}=a x^{2}\) (Neile's semicubical parabola)

A \(\$ 200\) investment in a savings account grows according to \(A(t)=200 e^{0.0398 t}\), for \(t \geq 0,\) where \(t\) is measured in years. a. Find the balance of the account after 10 years. b. How fast is the account growing (in dollars/year) at \(t=10 ?\) c. Use your answers to parts (a) and (b) to write the equation of the line tangent to the curve \(A=200 e^{0.0398 t}\) at the point \((10, A(10))\)

Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\) c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.

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