Question

Найдите тангенс угла наклона касательной, проведенной к графику функции f(x)=2x-x^2 в его точке с абциссой x0=-2

Answer 1

Тангенс угла равен значению 1 производной функции в нужной точке. y'=2-2*x⇒tgα=2-2*(-2)=2+4=6.

Ответ: 6.

Answer 2

Finding the Tangent of the Slope of the Tangent Line

To find the tangent of the slope of the tangent line to the graph of the function \( f(x) = 2x - x^2 \) at the point with the abscissa \( x_0 = -2 \), we can use the derivative of the function.

The derivative of the function \( f(x) = 2x - x^2 \) is \( f'(x) = 2 - 2x \).

Now, we can find the slope of the tangent line at the point \( x_0 = -2 \) by evaluating the derivative at that point: \[ f'(-2) = 2 - 2*(-2) = 2 + 4 = 6 \].

Therefore, the slope of the tangent line at the point \( x_0 = -2 \) is 6.

The tangent of the slope of the tangent line is simply the value of the slope itself, which is 6.