Consider the function f(x)=x−15x^1/3.

For the following questions, write inf for ∞, -inf for −∞, U for the union symbol, None if no answer exists, and separate by a comma if more than one answer exist.

a.) The domain of this function: (-inf,inf) .

b.) The x-intercept(s): ? and the y-intercept(s): 0
c.) The critical number(s): ? .

d.) f(x) is increasing on the interval(s): ? and decreasing on:? .

e.) f(x) has a local maximum at x =11.18035 and a local minimum at x = 11.180344.

f.) f(x) is concave up on the interval(s): ? and concave down on: ? .

g.) f(x) has one point of inflection at x = 0 .

h.) The Horizontal Asymptote is y = none and the vertical Asymptote is x = none .

a) The domain of the function f(x) = x - 15x^(1/3) is (-∞, ∞). To determine the domain, we need to consider any possible restrictions on the input values (x). In this case, there are no restrictions mentioned, so the domain includes all real numbers.

b) To find the x-intercepts, we set f(x) = 0 and solve for x.
x - 15x^(1/3) = 0
x^(2/3)(x^(1/3) - 15) = 0

Setting each factor equal to zero:
x^(2/3) = 0 --> x = 0

x^(1/3) - 15 = 0 --> x^(1/3) = 15 --> x = 15^3 = 3375

So, the function has two x-intercepts: x = 0 and x = 3375.

To find the y-intercept, we set x = 0 in the function:
f(0) = 0 - 15(0)^(1/3) = 0

Thus, the y-intercept is 0.

c) The critical numbers of a function occur where either the derivative is zero or does not exist. To find the critical numbers for f(x) = x - 15x^(1/3), we need to find its derivative, f'(x). Taking the derivative, we have:

f'(x) = 1 - 5x^(-2/3)

To find the critical number(s), we set f'(x) = 0 and solve for x:
1 - 5x^(-2/3) = 0
5x^(-2/3) = 1
x^(-2/3) = 1/5
(1/x)^(2/3) = 1/5^3
(1/x)^(2/3) = 1/125
(1/x) = (1/125)^(3/2)
x = (125^(3/2)) = 125 * 5^(3/2)

Therefore, the critical number is x ≈ 125 * 5^(3/2).

d) To determine where the function is increasing or decreasing, we need to examine the derivative. From part c), we know the derivative is:

f'(x) = 1 - 5x^(-2/3)

Since the sign of the derivative determines the function's behavior, we need to find where f'(x) > 0 and where f'(x) < 0.

For f'(x) > 0:
1 - 5x^(-2/3) > 0
1 > 5x^(-2/3)
x^(-2/3) < 1/5
(1/x)^(2/3) < 1/5
(1/x) < 1/5^(3/2)
x > 125 * 5^(3/2)

For f'(x) < 0:
1 - 5x^(-2/3) < 0
1 < 5x^(-2/3)
x^(-2/3) > 1/5
(1/x)^(2/3) > 1/5
(1/x) > 1/5^(3/2)
x < 125 * 5^(3/2)

Therefore, the function f(x) is increasing on the interval (125 * 5^(3/2), ∞) and decreasing on the interval (-∞, 125 * 5^(3/2)).

e) From the given information, f(x) has a local maximum at x = 11.18035 and a local minimum at x = 11.180344.

f) To determine the concavity of the function, we need to examine the second derivative. Taking the derivative again:

f''(x) = (1/3)(-2)(-5)x^(-2/3 - 1)
= (10/3)x^(-5/3)

Since the sign of the second derivative determines the concavity, we need to find where f''(x) > 0 and where f''(x) < 0.

For f''(x) > 0:
(10/3)x^(-5/3) > 0

Since the coefficient (10/3) is positive, the sign is determined by the base (x). Since x cannot be zero, the expression is always positive. Therefore, f''(x) > 0 for all x ≠ 0.

For f''(x) < 0:
(10/3)x^(-5/3) < 0

Since the coefficient (10/3) is positive, the sign is determined by the base (x). Since x cannot be zero, the expression is always negative. Therefore, f''(x) < 0 for all x ≠ 0.

Thus, the function f(x) is concave up on the interval (-∞, ∞) and concave down on the interval (-∞, ∞).

g) From the given information, f(x) has one point of inflection at x = 0.

h) The function f(x) = x - 15x^(1/3) does not have a horizontal asymptote, so the horizontal asymptote is y = None. Similarly, there is no value of x for which the function f(x) becomes infinitely large or undefined, so there is no vertical asymptote, and the vertical asymptote is x = None.

the function 3 9 10 3

x − x + , find the value of x for which the curve is i) rising ii) f