Example Question - function graph analysis
Here are examples of questions we've helped users solve.
Analysis of a Function Using Its Graph
<p>The domain of \( f \): All real numbers \( x \) such that \( -4 \leq x \leq 5 \).</p> <p>The range of \( f \): All real numbers \( y \) such that \( -3 \leq y \leq 5 \).</p> <p>The zeros of \( f \): \( x = -2, 1 \).</p> <p>\( f(-3.5) \): Cannot be determined exactly from the graph; not provided.</p> <p>The intervals on which \( f \) is increasing: \( (-4, -3) \cup (1, 5) \).</p> <p>The intervals on which \( f \) is decreasing: \( (-3, 1) \).</p> <p>The values for which \( f(x) < 0 \): \( x \) in intervals \( (-2, 1) \).</p> <p>Any relative maxima or minima: Relative maximum at \( x = -3 \), relative minimum at \( x = 1 \).</p> <p>The value(s) of \( x \) for which \( f(x) = 3 \): Approximately \( x = 4.5 \).</p> <p>Is \( f(0) \) positive or negative? Negative, since \( f(0) \) is below the \( x \)-axis.</p>
Limit Calculation on Function Graph
The image shows a function graphed on a coordinate plane, and there are several limits that need to be calculated. While I cannot interact with the image to enlarge it or click on it, I can analyze it based on the image provided. 1. The first limit is \(\lim_{x\to c^-} f(x)\), where `c` is the x-value at the vertical asymptote where the graph tends towards infinity as x approaches from the left. From the image, the graph shoots upwards towards positive infinity as x approaches this value from the left side. 2. The second limit is \(\lim_{x\to c^+} f(x)\), for the same value `c` (the vertical asymptote). From the graph, when approaching from the right side of the vertical asymptote, the function appears to head downwards towards negative infinity. 3. The third limit is \(\lim_{x\to d^-} f(x)\), where `d` represents the x-value where the function is heading toward as x approaches from the left. In this case, the graph approaches a horizontal asymptote from the left. However, without precise values or markings on the graph, I cannot indicate what the value of the limit is, so I can generally say that the limit has a certain finite value, which we can call 'L', where \(L\) is the y-value of the horizontal asymptote. 4. The fourth limit is \(\lim_{x\to d^+} f(x)\), which is as x approaches `d` from the right. From the graph, it seems the function approaches the same horizontal asymptote as it does from the left, thus also appearing to approach the same value 'L'. To summarize, the answers based on the graph (using general terms since specific values are not given): 1. \(\lim_{x\to c^-} f(x) = \infty\) 2. \(\lim_{x\to c^+} f(x) = -\infty\) 3. \(\lim_{x\to d^-} f(x) = L\) 4. \(\lim_{x\to d^+} f(x) = L\) Please bear in mind that 'L' stands for the finite limit value at the horizontal asymptote. In a precise solution, you would need to replace this with the actual y-coordinate value of the horizontal asymptote.
Analyzing Statements with a Decreasing Function Graph
The image shows a graph of a function f(x) which is decreasing on the interval [a, b]. The question asks which of the statements provided is incorrect given the graph. A) x · f'(x) < 0: Since x is positive on the interval [a, b] and f'(x) is negative (because the function is decreasing), their product is indeed negative. Therefore, this statement is correct. B) f(x) + f'(x) < 0: Since f(x) is positive (above the x-axis) and f'(x) is negative (the slope of the graph is negative), their sum might be negative or positive depending on the magnitudes. We can't definitively determine the sign of the sum without more information. This one cannot immediately be identified as false without further analysis, so we'll move on and come back if necessary. C) x - f'(x) > 0: We know x is positive and f'(x) is negative. Subtracting a negative number is equivalent to adding a positive number, so x - f'(x) would indeed be greater than 0. Therefore, this statement is correct. D) f(x) · -f'(x) > 0: Since f(x) is positive and -f'(x) is positive (because f'(x) is negative, and the negative of a negative is positive), their product will be positive. Therefore, this statement is correct. E) [f'(x)]^2 < 0: The square of any real number is non-negative. Since f'(x) is real, squaring it will give a non-negative result, so this statement is false. The incorrect statement is E) [f'(x)]^2 < 0.
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