Ask a New Question
Menu

Find the positive value of x where fx=x3+2x1/3 has a horizontal tangent line.

Answer =

1 answer

I will assume you meant:
f(x) = x^3 + 2x^(1/3)

then f'(x) = 3x^2 + (2/3)x^(-2/3)
for a horizontal tangent, f'(x) = 0
3x^2 + (2/3)/(x^(2/3)) = 0
multiply by x(2/3)
3x^(8/3) = -2/3
x^(8/3) = -2/9
x = (-2/9)^(3/8) , which is undefined (can't take an even root of a negative)

Unless you typed something different from what I assumed, there is no horizontal tangent.

see
http://www.wolframalpha.com/input/?i=x%5E3+%2B+2x%5E%281%2F3%29
Similar Questions
  1. Sorry but I've got a lot of problems that I don't understand.1) Let f(x)= (3x-1)e^x. For which value of x is the slope of the
    1. answers icon 3 answers
  2. find f'(x)find the slope of the graph of f a x=2 and x=4 find the equations of the tangent lines at x=2 and x=4 and find the
    1. answers icon 2 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  4. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
more similar questions