How do you find the equation of the line tangent to #y=x^3-x# at (0,0)?

Question

Elijah Abram · Accepted Answer

x + y = 0. The origin is a POI. The tangent crosses the curve. See all these features, in the Socratic graph.

#y'=3x^2-1, y''=6x and y'''=6 ne 0#. At, (0, 0), y'=-1, y''=0. So, the origin is a POI. The curve-crossing tangent at the POI (0, 0) is #y-0=-1(x-0)#, giving x + y = 0. graph{(y-x^3+x)(y+x)=0 [-10, 10, -5, 5]}

Christopher Agresta · Answer

undefined To find the equation of the line tangent to the curve y = x^3 - x at the point (0,0), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the function y = x^3 - x with respect to x.

The derivative of y = x^3 - x is given by dy/dx = 3x^2 - 1.

Substituting x = 0 into the derivative, we get dy/dx = 3(0)^2 - 1 = -1.

Therefore, the slope of the tangent line at (0,0) is -1.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Plugging in the values, we have y - 0 = -1(x - 0), which simplifies to y = -x.

Hence, the equation of the line tangent to y = x^3 - x at (0,0) is y = -x.