Question

F(x)=x^2+4x-7 найдите критические точки функции

Answer 1

Для нахождения критический точек приравниваем производную функции к нулю и решаем полученное уравнение:

$f'(x)=2x+4=0;\\2x=-4;\\x=-2.$

Ответ: x=-2 - единственная критическая точка

Answer 2

To find the critical points of a function, we need to determine where its derivative is equal to zero or undefined. Critical points are points on the graph where the function may have a local maximum, local minimum, or a horizontal inflection point.

First, let's find the derivative of the given function F(x):

F(x) = x^2 + 4x - 7

To find the derivative, apply the power rule:

F'(x) = d/dx(x^2) + d/dx(4x) - d/dx(7) F'(x) = 2x + 4

Now, set the derivative equal to zero and solve for x:

2x + 4 = 0 2x = -4 x = -4/2 x = -2

So, the critical point of the function is x = -2.