Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Let's break down each part of the problem step-by-step:
(A) Find all critical numbers of [tex]\( f \)[/tex].
To find the critical numbers, we first need to find the first derivative [tex]\( f'(x) \)[/tex] and then set it equal to zero and solve for [tex]\( x \)[/tex].
The first derivative of [tex]\( f(x) = 5x^6 - 2x^5 \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(5x^6 - 2x^5) = 30x^5 - 10x^4 = 10x^4(3x - 1) \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) = 0 \][/tex]
This implies [tex]\( x = 0 \)[/tex] or [tex]\( x = \frac{1}{3} \)[/tex].
So the critical numbers are:
[tex]\[ \text{Critical numbers } = [0, \frac{1}{3}] \][/tex]
(B) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is increasing.
To determine where [tex]\( f(x) \)[/tex] is increasing, we look at the intervals where the first derivative [tex]\( f'(x) \)[/tex] is positive.
Solving [tex]\( 30x^5 - 10x^4 > 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) > 0 \][/tex]
This inequality is satisfied when [tex]\( x > \frac{1}{3} \)[/tex], because [tex]\( x^4 \)[/tex] is always non-negative and is zero only at [tex]\( x = 0 \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is increasing in the interval:
[tex]\[ \text{Increasing: } (\frac{1}{3}, \infty) \][/tex]
(C) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is decreasing.
To determine where [tex]\( f(x) \)[/tex] is decreasing, we look at the intervals where the first derivative [tex]\( f'(x) \)[/tex] is negative.
Solving [tex]\( 30x^5 - 10x^4 < 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) < 0 \][/tex]
This inequality is satisfied when [tex]\( 0 < x < \frac{1}{3} \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is decreasing in the interval:
[tex]\[ \text{Decreasing: } (-\infty, 0) \cup (0, \frac{1}{3}) \][/tex]
(D) Find the [tex]\( x \)[/tex]-coordinates of all local maxima of [tex]\( f \)[/tex].
To find the local maxima, we check the second derivative [tex]\( f''(x) \)[/tex] at the critical points. If [tex]\( f''(x) < 0 \)[/tex] at a critical point, [tex]\( f(x) \)[/tex] has a local maximum there.
[tex]\[ f''(x) = \frac{d}{dx}(30x^5 - 10x^4) = 150x^4 - 40x^3 \][/tex]
[tex]\[ \text{Evaluating at } x = 0: \][/tex]
[tex]\[ f''(0) = 150(0)^4 - 40(0)^3 = 0 \][/tex]
[tex]\[ \text{Evaluating at } x = \frac{1}{3}: \][/tex]
[tex]\[ f''(\frac{1}{3}) = 150\left(\frac{1}{3}\right)^4 - 40\left(\frac{1}{3}\right)^3 = 150\left(\frac{1}{81}\right) - 40\left(\frac{1}{27}\right) = \frac{150}{81} - \frac{40}{27} = \frac{50}{27} - \frac{40}{27} = \frac{10}{27} > 0 \][/tex]
Thus, there are no local maxima:
[tex]\[ \text{x values of local maxima} = \text{NONE} \][/tex]
(E) Find the [tex]\( x \)[/tex]-coordinates of all local minima of [tex]\( f \)[/tex].
A local minimum occurs at a critical point where [tex]\( f''(x) > 0 \)[/tex].
From the calculation above, we find:
[tex]\[ f''(0) = 0 \][/tex]
[tex]\[ f''(\frac{1}{3}) = \frac{10}{27} > 0 \][/tex]
Thus, there is a local minimum at [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ \text{x values of local minima} = \frac{1}{3} \][/tex]
(F) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is concave up.
To determine concavity, we look at the second derivative [tex]\( f''(x) \)[/tex]. The function is concave up where [tex]\( f''(x) > 0 \)[/tex].
[tex]\[ 150x^4 - 40x^3 > 0 \][/tex]
Factoring out the common terms:
[tex]\[ 10x^3(15x - 4) > 0 \][/tex]
This inequality is satisfied when [tex]\( x < 0 \)[/tex] or [tex]\( x > \frac{4}{15} \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is concave up in the intervals:
[tex]\[ \text{Concave up: } (-\infty, 0) \cup (\frac{4}{15}, \infty) \][/tex]
So, let's compile everything:
(A) Critical numbers = [tex]\( [0, \frac{1}{3}] \)[/tex]
(B) Increasing: [tex]\( (\frac{1}{3}, \infty) \)[/tex]
(C) Decreasing: [tex]\( (-\infty, 0) \cup (0, \frac{1}{3}) \)[/tex]
(D) [tex]\( x \)[/tex] values of local maxima = NONE
(E) [tex]\( x \)[/tex] values of local minima = [tex]\( \frac{1}{3} \)[/tex]
(F) Concave up = [tex]\( (-\infty, 0) \cup (\frac{4}{15}, \infty) \)[/tex]
(A) Find all critical numbers of [tex]\( f \)[/tex].
To find the critical numbers, we first need to find the first derivative [tex]\( f'(x) \)[/tex] and then set it equal to zero and solve for [tex]\( x \)[/tex].
The first derivative of [tex]\( f(x) = 5x^6 - 2x^5 \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(5x^6 - 2x^5) = 30x^5 - 10x^4 = 10x^4(3x - 1) \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) = 0 \][/tex]
This implies [tex]\( x = 0 \)[/tex] or [tex]\( x = \frac{1}{3} \)[/tex].
So the critical numbers are:
[tex]\[ \text{Critical numbers } = [0, \frac{1}{3}] \][/tex]
(B) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is increasing.
To determine where [tex]\( f(x) \)[/tex] is increasing, we look at the intervals where the first derivative [tex]\( f'(x) \)[/tex] is positive.
Solving [tex]\( 30x^5 - 10x^4 > 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) > 0 \][/tex]
This inequality is satisfied when [tex]\( x > \frac{1}{3} \)[/tex], because [tex]\( x^4 \)[/tex] is always non-negative and is zero only at [tex]\( x = 0 \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is increasing in the interval:
[tex]\[ \text{Increasing: } (\frac{1}{3}, \infty) \][/tex]
(C) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is decreasing.
To determine where [tex]\( f(x) \)[/tex] is decreasing, we look at the intervals where the first derivative [tex]\( f'(x) \)[/tex] is negative.
Solving [tex]\( 30x^5 - 10x^4 < 0 \)[/tex]:
[tex]\[ 10x^4(3x - 1) < 0 \][/tex]
This inequality is satisfied when [tex]\( 0 < x < \frac{1}{3} \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is decreasing in the interval:
[tex]\[ \text{Decreasing: } (-\infty, 0) \cup (0, \frac{1}{3}) \][/tex]
(D) Find the [tex]\( x \)[/tex]-coordinates of all local maxima of [tex]\( f \)[/tex].
To find the local maxima, we check the second derivative [tex]\( f''(x) \)[/tex] at the critical points. If [tex]\( f''(x) < 0 \)[/tex] at a critical point, [tex]\( f(x) \)[/tex] has a local maximum there.
[tex]\[ f''(x) = \frac{d}{dx}(30x^5 - 10x^4) = 150x^4 - 40x^3 \][/tex]
[tex]\[ \text{Evaluating at } x = 0: \][/tex]
[tex]\[ f''(0) = 150(0)^4 - 40(0)^3 = 0 \][/tex]
[tex]\[ \text{Evaluating at } x = \frac{1}{3}: \][/tex]
[tex]\[ f''(\frac{1}{3}) = 150\left(\frac{1}{3}\right)^4 - 40\left(\frac{1}{3}\right)^3 = 150\left(\frac{1}{81}\right) - 40\left(\frac{1}{27}\right) = \frac{150}{81} - \frac{40}{27} = \frac{50}{27} - \frac{40}{27} = \frac{10}{27} > 0 \][/tex]
Thus, there are no local maxima:
[tex]\[ \text{x values of local maxima} = \text{NONE} \][/tex]
(E) Find the [tex]\( x \)[/tex]-coordinates of all local minima of [tex]\( f \)[/tex].
A local minimum occurs at a critical point where [tex]\( f''(x) > 0 \)[/tex].
From the calculation above, we find:
[tex]\[ f''(0) = 0 \][/tex]
[tex]\[ f''(\frac{1}{3}) = \frac{10}{27} > 0 \][/tex]
Thus, there is a local minimum at [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ \text{x values of local minima} = \frac{1}{3} \][/tex]
(F) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is concave up.
To determine concavity, we look at the second derivative [tex]\( f''(x) \)[/tex]. The function is concave up where [tex]\( f''(x) > 0 \)[/tex].
[tex]\[ 150x^4 - 40x^3 > 0 \][/tex]
Factoring out the common terms:
[tex]\[ 10x^3(15x - 4) > 0 \][/tex]
This inequality is satisfied when [tex]\( x < 0 \)[/tex] or [tex]\( x > \frac{4}{15} \)[/tex].
Thus, [tex]\( f(x) \)[/tex] is concave up in the intervals:
[tex]\[ \text{Concave up: } (-\infty, 0) \cup (\frac{4}{15}, \infty) \][/tex]
So, let's compile everything:
(A) Critical numbers = [tex]\( [0, \frac{1}{3}] \)[/tex]
(B) Increasing: [tex]\( (\frac{1}{3}, \infty) \)[/tex]
(C) Decreasing: [tex]\( (-\infty, 0) \cup (0, \frac{1}{3}) \)[/tex]
(D) [tex]\( x \)[/tex] values of local maxima = NONE
(E) [tex]\( x \)[/tex] values of local minima = [tex]\( \frac{1}{3} \)[/tex]
(F) Concave up = [tex]\( (-\infty, 0) \cup (\frac{4}{15}, \infty) \)[/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
