Given
f(x) = 2(x^(5/3)) - 5(x^(2/3))
Find the slope of the tangent at each x-intercept
To find the slope of the tangent at each x-intercept, we first need to find the x-intercepts of the function f(x). The x-intercepts are the values of x for which f(x) equals zero.
So, let's set f(x) equal to zero and solve for x:
2(x^(5/3)) - 5(x^(2/3)) = 0
Now, we can factor out a common term:
x^(2/3)(2x^(3/3) - 5) = 0
To find the x-intercepts, we set each factor equal to zero:
x^(2/3) = 0 or 2x^(3/3) - 5 = 0
For the first equation, x^(2/3) can only equal zero if x equals zero. So, we have one x-intercept at x = 0.
For the second equation, let's solve for x:
2x^(3/3) - 5 = 0
Simplifying,
2x - 5 = 0
Adding 5 to both sides,
2x = 5
Dividing by 2,
x = 5/2
Therefore, we have another x-intercept at x = 5/2.
Now that we have the x-intercepts, let's find the slopes of the tangent lines at each x-intercept. To find the slope at a given x-coordinate, we need to take the derivative of the function and evaluate it at that point.
The derivative of f(x) = 2(x^(5/3)) - 5(x^(2/3)) can be found using the power rule:
f'(x) = (2/3)(5/3)x^(5/3 - 1) - (2/3)(2/3)x^(2/3 - 1)
= (10/9)x^(2/3) - (4/9)x^(-1/3)
Now, let's evaluate the slopes at the x-intercepts.
At x = 0,
f'(0) = (10/9)(0)^(2/3) - (4/9)(0)^(-1/3)
= 0 - 0
= 0
So, the slope of the tangent line at x = 0 is 0.
At x = 5/2,
f'(5/2) = (10/9)(5/2)^(2/3) - (4/9)(5/2)^(-1/3)
To simplify this, we need to evaluate (5/2)^(2/3) and (5/2)^(-1/3):
(5/2)^(2/3) = (5/2)^(2/3) * (2/2)^(1/3)
= (10/2)^(2/3) * 2^(1/3)
= 10^(2/3) * 2^(1/3)
= (10 * 2)^(2/3)
= 20^(2/3)
= (20^(1/3))^2
= 2^2
= 4
(5/2)^(-1/3) = 1 / (5/2)^(1/3)
= 1 / [(5/2)^(1/3) * (2/2)^(2/3)]
= 1 / [(10/2)^(1/3) * 2^(2/3)]
= 1 / [10^(1/3) * 2^(2/3)]
= 1 / (10 * 2^(2/3))
= 1 / (10 * (2^(1/3))^2)
= 1 / (10 * 2^2)
= 1 / (10 * 4)
= 1 / 40
Now, let's substitute these values into the derivative:
f'(5/2) = (10/9)(4) - (4/9)(1/40)
= 40/9 - 1/90
= (40 * 10)/(9 * 10) - 1/90
= 400/90 - 1/90
= 399/90
So, the slope of the tangent line at x = 5/2 is 399/90.
Therefore, the slopes of the tangents at each x-intercept are 0 and 399/90.