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Magazine Chapter 2: Problem 47
Find the point(s), if any, at which the graph of has a horizontal tangent. $$ f(x)=\frac{x^{2}}{x-1} $$
Short Answer
Expert verified
The points at which the graph of \(f(x)=\frac{x^{2}}{x-1}\) has a horizontal tangent are \(x=0\) and \(x=2\).
Step by step solution
01
Obtain the Derivative of the function
Let's begin by finding the derivative of \(f(x)=\frac{x^{2}}{x-1}\). Since the function is given as quotient, use the quotient rule during differentiation. The quotient rule of differentiation states that the derivative of \(\frac{u}{v}\) is \(\frac{vu'-uv'}{v^{2}}\), where \(u' \) and \(v' \) are the derivatives of \(u \) and \(v \), respectively. Setting \(u=x^{2}\) and \(v=x-1\) yield \(u'=2x\) and \(v'=1\). Substituting into the rule gives us: \(f'(x)=\frac{(x-1)(2x)-x^{2}(1)}{(x-1)^{2}}=\frac{2x^{2}-2x-x^{2}}{(x-1)^{2}}=\frac{x^{2}-2x}{(x-1)^{2}}.\)
02
Set the Derivative Equal to 0
Set the derivative equal to 0 and solve for \(x\). Hence: \(0=\frac{x^{2}-2x}{(x-1)^{2}}\). A fraction is only zero if its numerator is zero. Therefore, the equation \(x^{2}-2x=0\) must be solved for \(x\).
03
Solve for x
Factoring \(x\) out of the equation \(x^{2}-2x=0\) gives \(x(x-2)=0\). Setting each factor equal to zero gives the solutions \(x=0\) and \(x=2\). However, since x=1 creates a singularity in the derivative, it should be checked if the limit around this point is 0. Since lim_{x->1} \(f'(x)\) does not equal to 0, there are no horizontal tangents at this point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, the derivative represents the rate of change of a function with respect to its variable. It's a measure of how a function's output value changes as the input changes. Derivatives are fundamental in finding slopes of tangent lines to curves on a graph, optimizing functions for maximum or minimum values, and solving many types of real-world problems.
For the function in our exercise, the derivative of \( f(x)=\frac{x^{2}}{x-1} \) tells us where the slope of the tangent to the curve is horizontal (slope = 0). To obtain this derivative, we use specific rules of differentiation such as the quotient rule, as the function is a ratio of two other functions.
For the function in our exercise, the derivative of \( f(x)=\frac{x^{2}}{x-1} \) tells us where the slope of the tangent to the curve is horizontal (slope = 0). To obtain this derivative, we use specific rules of differentiation such as the quotient rule, as the function is a ratio of two other functions.
Quotient Rule
The quotient rule is a procedure for differentiating a function that is formed as the quotient of two other functions. It states that if we have a function \( \frac{u}{v} \) , where both \( u \) and \( v \) are functions of \( x \) and are differentiable, the derivative of the quotient is given by \( \frac{vu' - uv'}{v^{2}} \), where \( u' \) and \( v' \) denote the derivatives of \( u \) and \( v \) respectively.
In the exercise, \( f(x) \) is the quotient of the functions \( x^2 \) and \( x-1 \) . Applying the quotient rule, we find that the derivative \( f'(x) = \frac{(x-1)(2x) - x^{2}(1)}{(x-1)^{2}} = \frac{x^{2}-2x}{(x-1)^{2}} \). This result will help us in identifying where the tangent line is horizontal.
In the exercise, \( f(x) \) is the quotient of the functions \( x^2 \) and \( x-1 \) . Applying the quotient rule, we find that the derivative \( f'(x) = \frac{(x-1)(2x) - x^{2}(1)}{(x-1)^{2}} = \frac{x^{2}-2x}{(x-1)^{2}} \). This result will help us in identifying where the tangent line is horizontal.
Critical Points
Critical points are values of \( x \) at which the derivative of a function either equals zero or does not exist. These points are important becau ... 2023-01-09 03:08:36.224 AppleWebKit/537.36 (KHTML, like Gecko) Chrome/97.0.4692.71 Safari/537.36 AppleWebKit/537.36 (KHTML, like Gecko) AppleWebKit/537.36 (KHTML, like Gecko) Safari/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 AppleWebKit/537.36 they help us determine where a function might have a maximum or minimum value, and where the graph of a function might have a horizontal tangent.
In our problem, we set the derivative equal to zero to find critical points, leading to the equation \( x^2 - 2x = 0 \) . Solving this, we find that the critical points are where \( x = 0 \) and \( x = 2 \) . It is important to note that the derivative does not exist at \( x = 1 \) , which is also critical information, as it indicates a potential vertical asymptote or point of discontinuity in the function, not a place where the tangent line could be horizontal.
In our problem, we set the derivative equal to zero to find critical points, leading to the equation \( x^2 - 2x = 0 \) . Solving this, we find that the critical points are where \( x = 0 \) and \( x = 2 \) . It is important to note that the derivative does not exist at \( x = 1 \) , which is also critical information, as it indicates a potential vertical asymptote or point of discontinuity in the function, not a place where the tangent line could be horizontal.
Factorization
Factorization is the process of breaking down an expression into a product of its factors. It is a powerful tool to simplify expressions and solve equations, particularly polynomial equations.
In the context of our problem, factorization simplifies the step of finding the critical points. For instance, \( x^2 - 2x = 0 \) can be factored as \( x(x - 2) = 0 \), where the solutions to this equation are found by setting each factor equal to zero. Thus, \( x = 0 \) and \( x = 2 \) are the points where the derivative is zero, suggesting potential locations for horizontal tangents on the graph of the function. Additionally, factorization assists in understanding the nature of the roots and can provide insight into the graph's symmetry and intercepts.
In the context of our problem, factorization simplifies the step of finding the critical points. For instance, \( x^2 - 2x = 0 \) can be factored as \( x(x - 2) = 0 \), where the solutions to this equation are found by setting each factor equal to zero. Thus, \( x = 0 \) and \( x = 2 \) are the points where the derivative is zero, suggesting potential locations for horizontal tangents on the graph of the function. Additionally, factorization assists in understanding the nature of the roots and can provide insight into the graph's symmetry and intercepts.
