Which is the graph of the cube root function f(x) = RootIndex 3 StartRoot x EndRoot?
1 answer:
The graph of a function f(x) = ∛x is shown in the picture, and the domain of the function will be all real numbers.
<h3>What is a function?</h3>It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
The question is incomplete.
The complete question is in the picture, please refer to the attached picture.
We have a function:
f(x) = ∛x
The domain of the function will be all real numbers.
x ∈(-∞, ∞)
The graph will be curved.
Plug x = 0, y = 0
x = 1, y = 1
The graph is shown in the attached picture.
Thus, the graph of a function f(x) = ∛x is shown in the picture, and the domain of the function will be all real numbers.
Learn more about the function here:
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Answer:
Step-by-step explanation:
A rectangular garden has vertices at
(x = 4, y = 3), (x = 6, y = 3), (x = 6, y = 9), and (x = 4, y = 9).
To plot each vertex of the garden find on the graph the x and y coordinate of the point and mark the point.
Answer:
See attached. Line for x<5.
Step-by-step explanation:
Solve the equation for x. Remember that the "<" and ">" signs are reversed when dividing or multiplying by a negative value.
9514 1404 393
Answer:
possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}
actual roots: -1, (2 ±4i√2)/3
no turning points; no local extrema
end behavior is same-sign as x-value end-behavior
Step-by-step explanation:
The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.
The Rational Root Theorem tells us any rational roots will be of the form ...
±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}
= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots
Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.
The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.
The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):
f(x) = (x +1)(3x^2 -4x +12)
The complex roots of the quadratic can be found using the quadratic formula:
x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6
x = (2 ± 4i√2)/3 . . . . complex roots
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The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.
