How do you find the points where the graph of the function #y = (x^3) + x# has horizontal tangents and what is the equation?

Question

Ezekiel Alexander · Accepted Answer

#y# has no real points through which there are horizontal tangents

#y=x^3+x# #y# would have horizontal tangents at points where #y'(x) =0# - if such points exist for #x in RR# #y'(x) = 3x^2+1# #y'(x) = 0 -> 3x^2+1=0# #x=sqrt(-1/3) = +-1/sqrt3i# Hence: #y'(x)=0# has no #x in RR# This can be seen by looking at the graph of #y# below: graph{x^3+x [-3.898, 3.896, -1.95, 1.948]}

Elijah Abram · Answer

undefined To find the points where the graph of the function y = (x^3) + x has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function y = (x^3) + x. Taking the derivative, we get: dy/dx = 3(x^2) + 1.

Next, we set the derivative equal to zero and solve for x: 3(x^2) + 1 = 0.

Simplifying the equation, we have: 3(x^2) = -1.

Dividing both sides by 3, we get: x^2 = -1/3.

Taking the square root of both sides, we have: x = ±√(-1/3).

Since we cannot take the square root of a negative number in the real number system, there are no real values of x that satisfy the equation x^2 = -1/3. Therefore, the graph of the function y = (x^3) + x does not have any points where it has horizontal tangents.