Answer :
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]
