Problem 75
Question
find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=\sqrt{x} $$
Step-by-Step Solution
Verified Answer
The difference quotient of the function \(f(x) = \sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\).
1Step 1: Substitute the function into the difference quotient formula
Initially, substitute \(f(x) = \sqrt{x}\) and \(f(x+h) = \sqrt{x+h}\) into the difference quotient. The formula becomes \(\frac{\sqrt{x+h}-\sqrt{x}}{h}\)
2Step 2: Simplify the equation
The expression in the numerator is a difference of squares, which cannot be simplified in its current form. To simplify it, multiply the top and bottom of the fraction by the conjugate of the numerator. The original expression thus becomes: \(\frac{\sqrt{x+h}-\sqrt{x}}{h} * \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\). Further simplifying, we get the equation to be \(\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\)
3Step 3: Simplify further and set the limit as h approaches 0
Now, cancel out the common factors from the numerator and the denominator, we get: \(\frac{1}{\sqrt{x+h}+\sqrt{x}}\). For the difference quotient, we need to take the limit as \(h\) goes to 0. Therefore, the difference quotient simplified to: \(\frac{1}{\sqrt{x+0}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\).
Key Concepts
Simplifying Rational ExpressionsRadicals and Rational ExponentsLimit of a Function
Simplifying Rational Expressions
Rational expressions are fractions where the numerator and denominator are polynomials. Simplifying them involves reducing them to their simplest form, making them easier to work with. In our example, we have a difference quotient, which is a rational expression because the numerator and denominator are each polynomial-like expressions once the square root is considered separately.
To simplify a rational expression, we usually look for common factors in the numerator and the denominator. But sometimes, like in this problem where we have radicals (square roots), it requires a bit of a different strategy. For instance, when working with the expression \(\frac{\sqrt{x+h}-\sqrt{x}}{h}\), multiplying the top and bottom by the conjugate is key to "rationalizing" the expression and opening up a simplification path.
Once you use the conjugate \(\sqrt{x+h}+\sqrt{x}\), you substitute back into the expression and simplify it. Pay close attention to eliminating terms strategically, as this helps to simplify complex rational expressions effectively.
To simplify a rational expression, we usually look for common factors in the numerator and the denominator. But sometimes, like in this problem where we have radicals (square roots), it requires a bit of a different strategy. For instance, when working with the expression \(\frac{\sqrt{x+h}-\sqrt{x}}{h}\), multiplying the top and bottom by the conjugate is key to "rationalizing" the expression and opening up a simplification path.
Once you use the conjugate \(\sqrt{x+h}+\sqrt{x}\), you substitute back into the expression and simplify it. Pay close attention to eliminating terms strategically, as this helps to simplify complex rational expressions effectively.
Radicals and Rational Exponents
Radicals, like \(\sqrt{x}\), can be rewritten using rational exponents. This is a powerful concept because it allows us to use all the rules of exponents to handle radicals more easily. For example, \(\sqrt{x}\) is the same as \(x^{1/2}\).
Understanding this relationship becomes crucial when simplifying expressions such as in a difference quotient involving radicals like \(\sqrt{x+h}\). It helps us manipulate the expressions better by applying exponent rules, although in this specific exercise, we stayed in the radical form for simplicity.
Understanding this relationship becomes crucial when simplifying expressions such as in a difference quotient involving radicals like \(\sqrt{x+h}\). It helps us manipulate the expressions better by applying exponent rules, although in this specific exercise, we stayed in the radical form for simplicity.
- Converting radicals to rational exponents: \(\sqrt[n]{x} = x^{1/n}\)
- Using exponent laws can simplify expressions and make derivative-related processes easier.
Limit of a Function
A limit describes the behavior of a function as the input approaches a certain value. It's foundational in calculus, especially for defining derivatives that include the difference quotient.
The limit of the difference quotient as \(h\) approaches 0 is pivotal to finding the derivative of a function. In our original problem, after simplifying the expression \(\frac{1}{\sqrt{x+h} + \sqrt{x}}\), we consider the limit as \(h\) tends to 0. This evaluates to \(\frac{1}{2\sqrt{x}}\), which represents the derivative of \(f(x) = \sqrt{x}\) at any given point.
The limit of the difference quotient as \(h\) approaches 0 is pivotal to finding the derivative of a function. In our original problem, after simplifying the expression \(\frac{1}{\sqrt{x+h} + \sqrt{x}}\), we consider the limit as \(h\) tends to 0. This evaluates to \(\frac{1}{2\sqrt{x}}\), which represents the derivative of \(f(x) = \sqrt{x}\) at any given point.
- Understanding limits enables precise calculation of function behavior at points that might otherwise result in undefined values, like dividing by zero.
- Limits deal with approximation, providing the accurate value as \(h\) gets extremely close to, but not equal to 0.
Other exercises in this chapter
Problem 74
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In Exercises \(75-82,\) express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=(3 x-1)^{4}$$
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Begin by graphing the square root function, \(f(x)=\sqrt{x} .\) Then use transformations of this graph to graph the given function. $$g(x)=\frac{1}{2} \sqrt{x+2
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