Question

Найти угол между касательной к графику функции Y=x^4-2x^3+3 В точке с абциссой X= 1 /2 и осью Ox

Answer 1

$y=x^{4}-2x^{3}+3$

Составим уравнение касательной в точке х=1/2

$y`=4x^{3}-6x^{2}=2x^{2}(2x-3)$

$y(\frac{1}{2})=(\frac{1}{2})^{4}-2(\frac{1}{2})^{3}+3=\frac{1}{16}-\frac{2}{8}+3=\frac{1-4+48}{16}=\frac{45}{16}$

$y`(\frac{1}{2})=2(\frac{1}{2})^{2}(2*\frac{1}{2}-3)=\frac{1}{2}(1-3)=\frac{1}{2}*(-2)=-1$

 

Уравнение касательной в общем виде:

$y=y(x_{0})+y`(x_{0})(x-x_{0})$

Подставим найденные значения в это уравнение:

$y=\frac{45}{16}+(-1)(x-\frac{1}{2})=\frac{45}{16}-x+\frac{1}{2}=\frac{53}{16}-x=-x+3\frac{5}{16}$

$y=-x+3\frac{5}{16}$ - уравнение касательной

 

Найдём угол между этой касательной и осью ОХ.

k=-1 (Коэффициент при х в уравнении касательной)

tga=k

tga=-1

a=135 градусов

 

 

Answer 2

Finding the Angle between the Tangent and the x-Axis

To find the angle between the tangent to the graph of the function Y=x^4-2x^3+3 at the point with abscissa X=1/2 and the x-axis, we can use the derivative of the function to find the slope of the tangent line at that point. Then, we can use the arctan function to find the angle.

The derivative of the function Y=x^4-2x^3+3 is given by: Y' = 4x^3 - 6x^2

Now, we can find the slope of the tangent line at the point where X=1/2 by substituting X=1/2 into the derivative.

The slope of the tangent line at X=1/2 is: Y'(1/2) = 4(1/2)^3 - 6(1/2)^2

Calculating the slope: Y'(1/2) = 1 - 3/2 = -1/2

The angle between the tangent and the x-axis can be found using the arctan function: Angle = arctan(-1/2)

Calculating the angle: Angle ≈ -26.57 degrees

So, the angle between the tangent to the graph of the function Y=x^4-2x^3+3 at the point with abscissa X=1/2 and the x-axis is approximately -26.57 degrees.