# How do you find the points where the graph of the function #y=2x^3# has horizontal tangents and what is the equation?

#y=0#

Given -

#y=2x^3#

It is a cubic function.

It has no constant terms. Hence it passes through the origin.

The slope of a horizontal tangent is

We have to find for which value of

The first derivative of the function gives the slope of the curve at any given point on the curve.

#dy/dx=6x^2#

#dy/dx=0 => 6x^2=0#

#x=0#

When

The tangent is through

Hence the equation of the tangent is

By signing up, you agree to our Terms of Service and Privacy Policy

To find the points where the graph of the function y=2x^3 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=2x^3. Taking the derivative, we get dy/dx = 6x^2.

Next, we set the derivative equal to zero and solve for x: 6x^2 = 0.

Solving this equation, we find that x = 0.

Therefore, the graph of the function y=2x^3 has horizontal tangents at the point (0,0).

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the derivative by definition for #y=x^(7/3)#?
- How do you estimate the instantaneous rate of change in population in 2010 if a city's population (tens of thousands) is modeled by the function #p(t) = 1.2(1.05)^t# where t is number of years since 2000?
- How do you find the slope of the secant lines of #f(x)=6.1x^2-9.1x# through the points: x=8 and x=16?
- What is the instantaneous velocity of an object moving in accordance to # f(t)= (e^(t^2),2t-te^t) # at # t=-1 #?
- How do you find the instantaneous rate of change from a table?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7