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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 67b

Let f(x) = 4√x - x.
Find all points on the graph of f at which the tangent line has slope -1/2.

Guida verificata passo dopo passo
1
Step 1: Find the derivative of the function f(x) = 4\(\sqrt{x}\) - x. The derivative, f'(x), represents the slope of the tangent line at any point x on the graph of f.
Step 2: Set the derivative equal to the given slope of the tangent line, which is -1/2. This will give you an equation to solve for x: f'(x) = -1/2.
Step 3: Solve the equation from Step 2 for x. This will give you the x-coordinates where the slope of the tangent line is -1/2.
Step 4: Substitute the x-values found in Step 3 back into the original function f(x) to find the corresponding y-coordinates. This will give you the points (x, y) on the graph.
Step 5: Verify the points by checking that the derivative at these x-values indeed equals -1/2, ensuring the solution is correct.

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Derivative

The derivative of a function measures the rate at which the function's value changes as its input changes. It is represented as f'(x) and provides the slope of the tangent line to the graph of the function at any given point. To find points where the tangent line has a specific slope, we need to compute the derivative and set it equal to that slope.
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Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. In this problem, we are looking for points on the graph of f where the slope of the tangent line equals -1/2.
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Percorso guidato
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Slopes of Tangent Lines

Finding Critical Points

Finding critical points involves determining where the derivative of a function is zero or undefined. These points are essential for analyzing the behavior of the function, including identifying where the slope of the tangent line meets specific criteria, such as -1/2 in this case. Solving the equation derived from the derivative will yield the x-values of interest.
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Critical Points