Skip to main content
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 37b

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = 1/ √x; a= 1/4

Guida verificata passo dopo passo
1
Step 1: Find the derivative of the function f(x) = \(\frac{1}{\sqrt{x}\)}. Rewrite the function as f(x) = x^{-1/2} to make differentiation easier.
Step 2: Use the power rule for differentiation, which states that \(\frac{d}{dx}\)[x^n] = nx^{n-1}, to find f'(x). For f(x) = x^{-1/2}, the derivative f'(x) = -\(\frac{1}{2}\)x^{-3/2}.
Step 3: Evaluate the derivative at the given point a = \(\frac{1}{4}\). Substitute x = \(\frac{1}{4}\) into f'(x) to find the slope of the tangent line at this point.
Step 4: Calculate f(a) by substituting a = \(\frac{1}{4}\) into the original function f(x) = \(\frac{1}{\sqrt{x}\)}. This will give you the y-coordinate of the point of tangency.
Step 5: Use the point-slope form of a line, y - y_1 = m(x - x_1), where m is the slope found in Step 3 and (x_1, y_1) is the point (\(\frac{1}{4}\), f(\(\frac{1}{4}\))), to write the equation of the tangent line.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
5m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is defined as the limit of the average rate of change of the function as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the graph of the function at that point.
Video consigliato:

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. The equation of the tangent line can be expressed using the point-slope form, which incorporates the slope and the coordinates of the point of tangency.
Video consigliato:
Percorso guidato
05:13
Slopes of Tangent Lines

Point-Slope Form

The point-slope form of a linear equation is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope (derivative) and the point of tangency are known. It allows for a straightforward way to express the line based on its slope and a specific point.
Video consigliato:
3:56
Slope-Intercept Form