Question

Найдите угловой коэффициент касательной к графику функции f(x)= x^3 - 3x^2 - 11 в точке с абсциссой x0=2.P.S. решение подробное.

Answer 1

По определению
$f'( x_{0} )=k$(угловому коэффициенту). Для начала найдем просто производную нашей функции
$f'(x)=3 x^{2} -6x$
Подставим значение $x_{0} =2$ в производную
$f'( x_{0} )=3* 2^{2} -6*2=0$
$k=0$

Answer 2

Finding the Slope of the Tangent Line

To find the slope of the tangent line to the graph of the function f(x) = x^3 - 3x^2 - 11 at the point with an x-coordinate of x0 = 2, we can use the derivative of the function. The derivative gives us the rate of change of the function at any given point.

The derivative of f(x) = x^3 - 3x^2 - 11 can be found by applying the power rule. The power rule states that if we have a function of the form f(x) = x^n, then the derivative is given by f'(x) = nx^(n-1).

Applying the power rule to the function f(x) = x^3 - 3x^2 - 11, we get:

f'(x) = 3x^2 - 6x

Now, to find the slope of the tangent line at x = 2, we substitute x = 2 into the derivative:

f'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0

Therefore, the slope of the tangent line to the graph of f(x) = x^3 - 3x^2 - 11 at the point with an x-coordinate of 2 is 0.

Solution:

The slope of the tangent line to the graph of the function f(x) = x^3 - 3x^2 - 11 at the point with an x-coordinate of x0 = 2 is 0.