Example Question - function evaluation
Here are examples of questions we've helped users solve.
Evaluating a Given Function Value from a Graph
<p>To find \( f(4) \), we locate the point where \( x = 4 \) on the graph and find the corresponding \( y \) value.</p> <p>At \( x = 4 \), the graph passes through the point \((4, 3)\).</p> <p>Therefore, \( f(4) = 3 \).</p>
Evaluating the Limit of a Function as It Approaches a Point
<p>The expression given is the definition of the derivative of \(\sqrt{x}\) evaluated at \(x=8\), which can be calculated as follows:</p> <p>\(\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) at \(x=8\)</p> <p>Let's apply the definition of the derivative for \(f(x) = \sqrt{x}\) at \(x = 8\):</p> <p>\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)</p> <p>\(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\)</p> <p>To evaluate the limit, multiply the numerator and the denominator by the conjugate of the numerator:</p> <p>\(= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\)</p> <p>Now plug in \(x = 8\):</p> <p>\(f'(8) = \lim_{h \to 0} \frac{1}{\sqrt{8+h} + \sqrt{8}}\)</p> <p>\(= \frac{1}{\sqrt{8} + \sqrt{8}}\)</p> <p>\(= \frac{1}{2\sqrt{8}}\)</p> <p>\(= \frac{1}{4\sqrt{2}}\)</p> <p>\(= \frac{\sqrt{2}}{8}\)</p> <p>Therefore, the limit and the derivative of \(\sqrt{x}\) at \(x = 8\) is \(\frac{\sqrt{2}}{8}\).</p>
Evaluating Limits of Functions
The question displays three functions and asks to evaluate the limits of certain expressions involving these functions. Given: - \( f: R \rightarrow R \) with \( f(x) = x^2 + 5x + 4 \) - \( g: R \rightarrow R \) with \( g(x) = 3x - 3 \) We are asked to find the following limits: a) \(\lim_{x \to -2} [f(x) + g(x)] \) b) \(\lim_{x \to 2} \sqrt{f(x)} - g(x) \) c) \(\lim_{x \to 3} \frac{f(x)}{g(x)} \) Let's solve each one: a) To find the limit of \( f(x) + g(x) \) as \( x \) approaches \(-2\), we simply plug \( -2 \) into each function and add the results: \( f(-2) = (-2)^2 + 5(-2) + 4 = 4 - 10 + 4 = -2 \) \( g(-2) = 3(-2) - 3 = -6 - 3 = -9 \) Now, \( f(-2) + g(-2) = -2 + (-9) = -11 \) So, \(\lim_{x \to -2} [f(x) + g(x)] = -11\). b) To find the limit of \( \sqrt{f(x)} - g(x)\) as \(x\) approaches \(2\), we substitute \(2\) into each function and then perform the operations, keeping in mind to take the square root of \(f(x)\) not \(f(x)\) itself: \( f(2) = (2)^2 + 5(2) + 4 = 4 + 10 + 4 = 18 \) \( g(2) = 3(2) - 3 = 6 - 3 = 3 \) Now, the square root of \( f(2) \) is \( \sqrt{18} \), so: \( \sqrt{f(2)} - g(2) = \sqrt{18} - 3 \) This result cannot be simplified further without a calculator, so we leave it as \( \sqrt{18} - 3 \). Therefore, \(\lim_{x \to 2} \sqrt{f(x)} - g(x) = \sqrt{18} - 3\). c) To find the limit of \( \frac{f(x)}{g(x)}\) as \(x\) approaches \(3\), we again substitute \(3\) into each function and divide the results: \( f(3) = (3)^2 + 5(3) + 4 = 9 + 15 + 4 = 28 \) \( g(3) = 3(3) - 3 = 9 - 3 = 6 \) Now, \( \frac{f(3)}{g(3)} = \frac{28}{6} = \frac{14}{3} \) So, \(\lim_{x \to 3} \frac{f(x)}{g(x)} = \frac{14}{3}\). To summarize, the limits are as follows: a) \(-11\) b) \( \sqrt{18} - 3 \) c) \( \frac{14}{3} \)
Calculating a Function Value
The image shows a function definition and a request to evaluate the function at a particular point: f(x) = x^2 - x; f(-4) To solve this, we simply substitute x with -4 in the function's equation: f(-4) = (-4)^2 - (-4) f(-4) = 16 + 4 f(-4) = 20 Therefore, f(-4) is equal to 20.
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