Question

F(x)=4x+sin3x, x0=-π/2​

Answer 1

Відповідь:

f(0)=4*0+sin(3*0)

f(0)=4*0+sin(3*0)

f(0)=0+sin(0)

Покрокове пояснення:

Answer 2

It looks like you want to find the value of the function $F(x) = 4x + \sin(3x)$ at the point $x_0 = -\frac{\pi}{2}$. To do this, you'll substitute $x_0$ into the function and calculate the result. Let's go through the steps:

Given function: $F(x) = 4x + \sin(3x)$ Point: $x_0 = -\frac{\pi}{2}$

Substitute $x_0$ into the function: $F\left(-\frac{\pi}{2}\right) = 4\left(-\frac{\pi}{2}\right) + \sin\left(3 \cdot \left(-\frac{\pi}{2}\right)\right)$

Now, calculate the value: $F\left(-\frac{\pi}{2}\right) = -2\pi + \sin\left(-\frac{3\pi}{2}\right)$

Since $\sin$ has a periodicity of $2\pi$, you can simplify the argument: $\sin\left(-\frac{3\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1$

Now, plug this back into the expression: $F\left(-\frac{\pi}{2}\right) = -2\pi - 1$

So, the value of the function $F(x)$ at $x_0 = -\frac{\pi}{2}$ is $-2\pi - 1$.