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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 3 - Derivatives문제 3.2.56
3장, 문제 3.2.56

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?

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To find the tangent line to the curve y = √x, we first need to find the derivative of y with respect to x. The derivative, y', represents the slope of the tangent line at any point x. For y = √x, the derivative is y' = (1/2)x^(-1/2).
Next, we need to determine the equation of the tangent line. The equation of a line in point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point of tangency. We will use the derivative to find the slope m at a specific point x₁.
To check if any tangent line crosses the x-axis at x = -1, we set y = 0 in the tangent line equation and solve for x. This gives us the x-intercept of the tangent line. Substitute y = 0 into the equation y - y₁ = m(x - x₁) and solve for x.
Substitute the expression for the slope m = (1/2)x₁^(-1/2) and the point of tangency (x₁, √x₁) into the equation of the tangent line. This gives us y - √x₁ = (1/2)x₁^(-1/2)(x - x₁).
Finally, solve the equation 0 - √x₁ = (1/2)x₁^(-1/2)(-1 - x₁) to check if there exists a value of x₁ such that the tangent line crosses the x-axis at x = -1. If a solution exists, find the corresponding x₁ and y₁ to determine the point of tangency and the equation of the tangent line.

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주요 개념

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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. For the function y = √x, finding the tangent line involves calculating the derivative and using the point-slope form of a line.
추천 영상:
가이드 코스
05:13
Slopes of Tangent Lines

Derivative

The derivative of a function measures how the function's output changes as its input changes. For y = √x, the derivative is found using the power rule, resulting in dy/dx = 1/(2√x). This derivative provides the slope of the tangent line at any point on the curve, which is essential for determining the equation of the tangent line.
추천 영상:
가이드 코스
05:44
Derivatives

Intersection with the x-axis

A line intersects the x-axis when its y-coordinate is zero. To find if a tangent line crosses the x-axis at x = -1, we set the equation of the tangent line to zero and solve for x. Since the function y = √x is only defined for x ≥ 0, any tangent line derived from this function cannot cross the x-axis at x = -1, as it would imply a point of tangency in the undefined region.
추천 영상:
06:30
Disk Method Using y-Axis
관련 실천
교과서 질문

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.

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교과서 질문

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)

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교과서 질문

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x³ + y³ = 18xy

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교과서 질문

Derivatives


In Exercises 1–18, find dy/dx.


f(x) = x³ sin x cos x

301
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교과서 질문

Finding Derivative Values


In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.


f(u) = 1 − (1/u), u = g(x) = (1 / (1 − x)), x = −1

243
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교과서 질문

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)

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