Can someone help me finish this question... I got most of the answers but need help.
Suppose that f(x)=3x−6x+5.
(A) Use interval notation to indicate where f(x) is defined. If it is defined on more than one interval, enter the union of all intervals where f(x) is defined.
Domain: (-inf,-5)U(-5,inf)
(B) Find all intercepts. If there are no intercepts, enter None. If there are more than one, enter them separated by commas.
x-intercept(s) = 2
y-intercept(s) = ?
(C) Find all critical values of f. If there are no critical values, enter None. If there are more than one, enter them separated by commas.
Critical value(s) = none.
(D) Use interval notation to indicate where f(x) is increasing or decreasing. If there are more than one interval, enter the union of all intervals. If the answer is the empty set, enter {}.
Increasing: ?
Decreasing: ?
(E) Find the x-coordinates of all local maxima and minima of f. If there are no local maxima, enter None. If there are more than one, enter them separated by commas.
Local maxima at x = none.
Local minima at x = none.
(F) Use interval notation to indicate where f(x) is concave up or down.
Concave up: ?
Concave down: ?
(G) Find all inflection points of f. If there are no inflection points, enter None. If there are more than one, enter them separated by commas.
Inflection point(s) at x = ?
(H) Find all asymptotes of f. If there are no asymptotes, enter None. If there are more than one, enter them separated by commas.
Horizontal asymptote(s): y = 3
Vertical asymptote(s): x = -5
Assuming you really meant to type
f(x) = 3x^2 - 6x + 5
then
f'(x) = 6x - 6
and
f''(x) = 6
All polynomials are defined for all real values of x.
x-intercepts: where f(x) = 0
y-intercept: f(x) where x=0
max/min where f'(x) = 0 and f''(x) is not zero
increasing where f'(x) > 0
concave up where f''(x) > 0
inflection where f''(x) = 0
Polynomials have no asymptotes. Rational functions usually have asymptotes
So, having reviewed the information above, what are your answers? We'll be happy to check them for you.
(B) To find the y-intercept, we need to determine the value of f(x) when x=0. Substitute x=0 into the function f(x)=3x-6x+5:
f(0) = 3(0) - 6(0) + 5 = 5
So the y-intercept is 5.
(D) To determine where f(x) is increasing or decreasing, we need to find the intervals where the function is either increasing or decreasing. First, we can rewrite the function f(x):
f(x) = 3x - 6x + 5 = -3x + 5
Now, we can find where the derivative of f(x) is positive (indicating the function is increasing) or negative (indicating the function is decreasing). Taking the derivative of f(x) gives:
f'(x) = -3
Since f'(x) is a constant (-3), it is always negative. This means that f(x) is always decreasing for all values of x.
So the answer for (D) is:
Increasing: {}
Decreasing: (-inf, inf)
(E) To find the local maxima and minima of f(x), we need to find the critical values. Critical values are the values of x where the derivative is either zero or undefined.
Taking the derivative of f(x) gives us: f'(x) = -3
Since the derivative is a constant, there are no critical values. Therefore, there are no local maxima or minima.
So the answer for (E) is:
Local maxima at x = none.
Local minima at x = none.
(F) To determine where f(x) is concave up or down, we need to find the second derivative of f(x). Taking the derivative of f'(x) gives us:
f''(x) = 0
Since the second derivative is zero, it does not indicate whether the function is concave up or down.
So the answer for (F) is:
Concave up: {}
Concave down: {}
(G) To find the inflection points of f(x), we need to find the x-values where the concavity changes. In this case, since the second derivative is zero, there are no inflection points.
So the answer for (G) is:
Inflection point(s) at x = none.
(H) To find the asymptotes of f(x), we need to examine the behavior of the function as x approaches certain values.
- There is a horizontal asymptote at y=3, as x approaches positive or negative infinity, the function tends towards 3.
- There is a vertical asymptote at x=-5, as x approaches -5, the function tends towards infinity or negative infinity.
So the answer for (H) is:
Horizontal asymptote(s): y = 3
Vertical asymptote(s): x = -5