Then, find the maximum and global maximum on the interval -1 <= x <= 7.
I'm still confused on taking the derivative of a fractional exponent.
I'm still confused on taking the derivative of a fractional exponent.
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To take the derivative of a rational (fractional) exponent, the same rules apply as for whole numbers:
Let g(x) = xⁿ, where n is any real number.
Then g'(x) = n xⁿ⁻¹
For a rational exponent, such as 2/5, you simply write 2/5 as a coefficient of the function, and subtract 1 from 2/5:
Let h(x) = (x - 2)^(2/5)
Then h'(x) = (2/5) (x - 2)^(2/5 - 1)
h'(x) = 2/5 (x - 2)^(2/5 - 5/5)
h'(x) = 2/5 (x - 2)^(-3/5)
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f(x) = 3x^(1/3) - x
f '(x) = 3(1/3) x^(-2/3) - 1
f '(x) = x^(-2/3) - 1
Critical points occur when f '(x) is zero or undefined.
= = = = =
f '(x) = 0
= = = = =
f '(x) = 1/x^(2/3) - 1 = 0
1/x^(2/3) = 1
1 = x^(2/3)
1 = ³√(x²)
(1)³ = (³√(x²))³
1 = √(x²)
1 = |x|
x = -1, 1
f '(x) = undefined when x = 0
Critical points occur when x = -1, x = 0, and x = 1.
Apply the first derivative test:
On the interval (-1, 0), f '(x) is negative. For example, f '(-1/2) = 1/(³√4) - 1 ≈ -0.37. Therefore, f is decreasing on this interval.
On (0, 1), f '(x) > 0, so f is increasing.
On (1, 7), f '(x) < 0, so f is decreasing.
Although there is a decrease-increase pattern before and after x = 0, there is no relative extremum here because f(x) is undefined and discontinuous at x = 0.
When x = 1, there is a local maximum, and this max occurs at the point (1, 2).
To check for the global maximum on the interval [-1, 7], you also have to take into account the endpoints.
f(-1) = -2
f(7) ≈ -1.26
And so, the global max on the interval is (1, 2).
Let g(x) = xⁿ, where n is any real number.
Then g'(x) = n xⁿ⁻¹
For a rational exponent, such as 2/5, you simply write 2/5 as a coefficient of the function, and subtract 1 from 2/5:
Let h(x) = (x - 2)^(2/5)
Then h'(x) = (2/5) (x - 2)^(2/5 - 1)
h'(x) = 2/5 (x - 2)^(2/5 - 5/5)
h'(x) = 2/5 (x - 2)^(-3/5)
- - - - - - - - - - - - - - -
f(x) = 3x^(1/3) - x
f '(x) = 3(1/3) x^(-2/3) - 1
f '(x) = x^(-2/3) - 1
Critical points occur when f '(x) is zero or undefined.
= = = = =
f '(x) = 0
= = = = =
f '(x) = 1/x^(2/3) - 1 = 0
1/x^(2/3) = 1
1 = x^(2/3)
1 = ³√(x²)
(1)³ = (³√(x²))³
1 = √(x²)
1 = |x|
x = -1, 1
f '(x) = undefined when x = 0
Critical points occur when x = -1, x = 0, and x = 1.
Apply the first derivative test:
On the interval (-1, 0), f '(x) is negative. For example, f '(-1/2) = 1/(³√4) - 1 ≈ -0.37. Therefore, f is decreasing on this interval.
On (0, 1), f '(x) > 0, so f is increasing.
On (1, 7), f '(x) < 0, so f is decreasing.
Although there is a decrease-increase pattern before and after x = 0, there is no relative extremum here because f(x) is undefined and discontinuous at x = 0.
When x = 1, there is a local maximum, and this max occurs at the point (1, 2).
To check for the global maximum on the interval [-1, 7], you also have to take into account the endpoints.
f(-1) = -2
f(7) ≈ -1.26
And so, the global max on the interval is (1, 2).