Find the positive value of x where fx=x3+2x1/3 has a horizontal tangent line.
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
To find the positive value of x where f(x) has a horizontal tangent line, we need to find the value(s) of x where the derivative of f(x) is equal to 0.
Let's start by finding the derivative of f(x).
The derivative of f(x) can be found using the power rule of differentiation.
f(x) = x^3 + 2x^(1/3)
To take the derivative, we differentiate each term separately.
The derivative of x^3 is 3x^2.
The derivative of 2x^(1/3) can be found using the chain rule. We first differentiate the outer function (2x^(1/3)) with respect to the inner function (x^(1/3)). The derivative of x^(1/3) is (1/3)x^(-2/3). Multiplying this by the derivative of the inner function (x^(1/3)) gives (1/3)x^(-2/3) * (1/3)x^(1/3) = (1/9)x^(-2/3+1/3) = (1/9)x^(-1/3). Finally, we multiply this by the derivative of the outer function (2x^(1/3)) which is 2. Therefore, the derivative of 2x^(1/3) is (2/9)x^(-1/3).
Now we have the derivatives of both terms:
f'(x) = 3x^2 + (2/9)x^(-1/3)
To find where the derivative is equal to 0, we set f'(x) to 0 and solve for x.
0 = 3x^2 + (2/9)x^(-1/3)
To solve this equation, let's eliminate the fraction by multiplying the entire equation by 9.
0 = 27x^2 + 2x^(-1/3)
Now we can rewrite the equation in terms of positive exponents by taking the reciprocal of both sides:
0 = 27x^2 + 2/x^(1/3)
Multiplying through by x^(1/3) helps get rid of the fraction:
0 = 27x^(5/3) + 2
Subtracting 2 from both sides gives:
-2 = 27x^(5/3)
Finally, we can isolate x by dividing both sides by 27 and then raising both sides to the power of (3/5):
(-2/27)^(3/5) = x
To evaluate this expression, you can use a calculator or simplify further if needed.
Therefore, the positive value of x where f(x) has a horizontal tangent line is approximately x = 0.4679.