Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.1.5

Capitolo 3, Problema 3.1.5

In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.

y = 4 − x², (−1, 3)

Guida verificata passo dopo passo

First, identify the function given: \( y = 4 - x^2 \). This is a quadratic function representing a parabola that opens downwards.

To find the equation of the tangent line, we need the derivative of the function, which gives us the slope of the tangent line at any point \( x \). Differentiate \( y = 4 - x^2 \) with respect to \( x \) to get \( \frac{dy}{dx} = -2x \).

Evaluate the derivative at the given point \( x = -1 \) to find the slope of the tangent line. Substitute \( x = -1 \) into \( \frac{dy}{dx} = -2x \) to get the slope \( m = -2(-1) = 2 \).

Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-1, 3) \) and \( m = 2 \). Substitute these values into the equation to get \( y - 3 = 2(x + 1) \).

Simplify the equation \( y - 3 = 2(x + 1) \) to get the final equation of the tangent line. This will give you the equation in the form \( y = mx + b \), which can be used to sketch the tangent line along with the curve \( y = 4 - x^2 \).

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:

Concetti chiave

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

Derivative

The derivative of a function at a point provides the slope of the tangent line to the curve at that point. For the function y = 4 - x², the derivative is found using basic differentiation rules, resulting in dy/dx = -2x. Evaluating this derivative at x = -1 gives the slope of the tangent line at the point (-1, 3).

Video consigliato:

Percorso guidato

05:44

Derivatives

Point-Slope Form of a Line

The point-slope form is a method for writing the equation of a line when you know a point on the line and its slope. It is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point. Using the slope from the derivative and the point (-1, 3), you can find the equation of the tangent line.

Video consigliato:

Percorso guidato

05:13

Slopes of Tangent Lines

Graphing Functions and Tangent Lines

Graphing involves plotting the curve of the function and the tangent line to visualize their relationship. For y = 4 - x², the graph is a downward-opening parabola. The tangent line at (-1, 3) will touch the curve at this point, illustrating the concept of tangency where the line just 'kisses' the curve without crossing it.

Video consigliato:

Percorso guidato

05:13

Slopes of Tangent Lines

Pratica correlata

Domanda del libro di testo

Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = πh³. The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h.

236

views

Domanda del libro di testo

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

f(x) = √(x + 1), (8, 3)

229

views

Domanda del libro di testo

Tangent line to y = √x Does any tangent line to the curve y = √x cross the x-axis at x = −1? If so, find an equation for the line and the point of tangency. If not, why not?

240

views

Domanda del libro di testo

Derivatives

In Exercises 1–18, find dy/dx.

f(x) = x³ sin x cos x

301

views

Domanda del libro di testo

Derivatives

In Exercises 27–32, find dp/dq.

p = (q sin q) / (q² − 1)