Logo

AskSia

Plus

Find the slope of the tangent to the curve y=3x2−2x+1y = 3x^2 - 2x + 1 at x=1x = 1.
Jul 18, 2024
Find the slope of the tangent to the curve y=3x2−2x+1y = 3x^2 - 2x + 1 at x=1x = 1.
Generated Graph
Solution by Steps
step 1
To find the slope of the tangent line to the curve y=3x2−2x+1y = 3x^2 - 2x + 1 at a specific point, we need to compute the derivative of yy with respect to xx
step 2
The derivative of yy with respect to xx is y′=dydx=6x−2y' = \frac{dy}{dx} = 6x - 2
step 3
To find the slope at x=1x = 1, we substitute 11 into the derivative: y′(1)=6(1)−2y'(1) = 6(1) - 2
step 4
Simplifying gives us the slope at x=1x = 1: y′(1)=6−2=4y'(1) = 6 - 2 = 4
Answer
The slope of the tangent to the curve at x=1x = 1 is 44.
Key Concept
Finding the slope of a tangent line at a point
Explanation
The slope of the tangent line to a curve at a given point is the value of the derivative of the curve's equation evaluated at that point.
解方程 x3−3x2+3x−1=0 x^3 - 3x^2 + 3x - 1 = 0 并验证找到的根
Generated Graph
Solution by Steps
step 1
Identify the given polynomial equation to solve: x3−3x2+3x−1=0 x^3 - 3x^2 + 3x - 1 = 0
step 2
Recognize that the polynomial resembles the expansion of (x−1)3 (x - 1)^3
step 3
Factor the polynomial as (x−1)3=0 (x - 1)^3 = 0
step 4
Solve the equation (x−1)3=0 (x - 1)^3 = 0 by taking the cube root of both sides to find x=1 x = 1
step 5
Verify the root by substituting x=1 x = 1 into the original equation: 13−3(1)2+3(1)−1 1^3 - 3(1)^2 + 3(1) - 1
step 6
Simplify the substitution to confirm the root: 1−3+3−1=0 1 - 3 + 3 - 1 = 0
Answer
The root of the equation x3−3x2+3x−1=0 x^3 - 3x^2 + 3x - 1 = 0 is x=1 x = 1 .
Key Concept
Factoring Polynomials
Explanation
The polynomial x3−3x2+3x−1 x^3 - 3x^2 + 3x - 1 can be factored as (x−1)3 (x - 1)^3 because it resembles the binomial expansion of (x−1)3 (x - 1)^3 . The root of the equation is found by setting the factored form equal to zero and solving for x x .
计算定积分 ∫01ln(x)xdx \int_{0}^{1} \frac{ln(x)}{x} dx
Generated Graph
Solution by Steps
step 1
Evaluate the integral ∫01ln⁡(x)xdx\int_{0}^{1} \frac{\ln(x)}{x} dx
step 2
Recognize that the integral of ln⁡(x)x\frac{\ln(x)}{x} from 00 to 11 does not converge
Answer
The integral ∫01ln⁡(x)xdx\int_{0}^{1} \frac{\ln(x)}{x} dx does not converge.
Key Concept
Convergence of Improper Integrals
Explanation
The integral of ln⁡(x)x\frac{\ln(x)}{x} from 00 to 11 is an improper integral because ln⁡(x)x\frac{\ln(x)}{x} is not defined at x=0x=0. When evaluating such an integral, one must check if the limit exists as the lower bound approaches 00. In this case, the limit does not exist, and therefore, the integral does not converge.
What is the result of the definite integral ∫01ln⁡(x)xdx \int_{0}^{1} \frac{\ln(x)}{x} dx ?
Generated Graph
Solution by Steps
step 1
Evaluate the definite integral ∫01ln⁡(x)xdx \int_{0}^{1} \frac{\ln(x)}{x} dx
step 2
Recognize that the integral involves the natural logarithm function divided by xx
step 3
Determine the convergence of the integral
step 4
Conclude that the integral does not converge because the integrand ln⁡(x)x\frac{\ln(x)}{x} is not defined at x=0x = 0
Answer
The integral ∫01ln⁡(x)xdx \int_{0}^{1} \frac{\ln(x)}{x} dx does not converge.
Key Concept
Convergence of Improper Integrals
Explanation
The integral of ln⁡(x)x\frac{\ln(x)}{x} from 00 to 11 is an improper integral because the integrand becomes infinite as xx approaches 00. Such integrals require careful analysis to determine if they converge or diverge. In this case, the integral does not converge.
Generate me a similar question
What is the equation of the tangent line to the curve y=2x2+3x−1y = 2x^2 + 3x - 1 at the point x=2x = 2?
© 2023 AskSia.AI all rights reserved