Question

Найдите дельта х и дельта f в точке х01)f(x)=pi/3+sin x; x0=2pi/3; x=3pi/42)f(x)=ctg x - pi/4; x0=pi/4; x=pi/3НУЖНО ПОЛНОЕ РЕШЕНИЕ

Answer 1

1)$x_{0}=\frac{2\pi}{3}\\ x=(x_{0}+\Delta x)=\frac{3\pi}{4}\\ \Delta x=x-x_{0}=\frac{3\pi}{4}-\frac{2\pi}{3}=\frac{9\pi-8\pi}{12}=\frac{\pi}{12}\\ \Delta y=f(x_{0}+\Delta x)-f(x_{0})\\ \Delta y=(\frac{\pi}{3}+sin\frac{3\pi}{4})-(\frac{\pi}{3}+sin\frac{2\pi}{3})=sin\frac{3\pi}{4}-sin\frac{2\pi}{3}=\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}=\\ =\frac{\sqrt{2}-\sqrt{3}}{2}$

 

2)$x_{0}=\frac{\pi}{4}\\ x=(x_{0} +\Delta x) = \frac{\pi}{3}\\ \Delta x = x-x_{0}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\\ \Delta y=f(x_{0}+\Delta x) -f(x_{0})\\ \Delta y =(ctg\frac{\pi}{3}-\frac{\pi}{4})-(ctg\frac{\pi}{4}-\frac{\pi}{4})=ctg\frac{\pi}{3}-ctg\frac{\pi}{4}=\frac{1}{\sqrt{3}} -1=\frac{\sqrt{3}-3}{3}$

Answer 2

Finding Δx and Δf in point x₀ for f(x) = π/3 + sin(x); x₀ = 2π/3; x = 3π/4

To find Δx and Δf in point x₀ for the function f(x) = π/3 + sin(x), we need to calculate the change in x (Δx) and the change in f (Δf) between x₀ and x.

First, let's calculate Δx: Δx = x - x₀ = 3π/4 - 2π/3

To simplify the calculation, we can find a common denominator for 4 and 3, which is 12: Δx = (9π/12) - (8π/12) = π/12

Next, let's calculate Δf: Δf = f(x) - f(x₀) = (π/3 + sin(3π/4)) - (π/3 + sin(2π/3))

Using trigonometric identities, we can simplify the expression: Δf = (π/3 + √2/2) - (π/3 + √3/2) Δf = √2/2 - √3/2

Therefore, in point x₀ for f(x) = π/3 + sin(x), Δx = π/12 and Δf = √2/2 - √3/2.

Finding Δx and Δf in point x₀ for f(x) = ctg(x) - π/4; x₀ = π/4; x = π/3

To find Δx and Δf in point x₀ for the function f(x) = ctg(x) - π/4, we need to calculate the change in x (Δx) and the change in f (Δf) between x₀ and x.

First, let's calculate Δx: Δx = x - x₀ = π/3 - π/4

To simplify the calculation, we can find a common denominator for 3 and 4, which is 12: Δx = (4π/12) - (3π/12) = π/12

Next, let's calculate Δf: Δf = f(x) - f(x₀) = (ctg(π/3) - π/4) - (ctg(π/4) - π/4)

Using trigonometric identities, we can simplify the expression: Δf = (√3 - π/4) - (1 - π/4) Δf = √3 - 1

Therefore, in point x₀ for f(x) = ctg(x) - π/4, Δx = π/12 and Δf = √3 - 1.

I hope this helps! Let me know if you have any further questions.