Respuesta :
Answers:
(a) f is increasing at [tex](-\infty,-2) \cup (2,\infty)[/tex].
(b) f is decreasing at [tex](-2,2)[/tex].
(c) f is concave up at [tex](2, \infty)[/tex]
(d) f is concave down at [tex](-\infty, 2)[/tex]
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
[tex]f'(x) = 15x^2 - 60 [/tex]
So,
[tex] f'(x) \ \textgreater \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}[/tex]
The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:
-->> x < -2
-->> -2 < x < 2
--->> x > 2
If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.
If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.
So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since
[tex]f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0[/tex]
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at [tex](-\infty,-2) \cup (2,\infty)[/tex].
(b) f is decreasing only when the derivative of f is negative. Since
[tex]f'(x) = 15x^2 - 60 [/tex]
Using the similar computation in (a),
[tex]f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}[/tex]
Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.
Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)
(c) f is concave up if and only if the second derivative of f is positive. Note that
[tex]f''(x) = 30x - 60[/tex]
Since,
[tex]f''(x) \ \textgreater \ 0 \\ \\ \Leftrightarrow 30x - 60 \ \textgreater \ 0 \\ \\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0 \\ \\ \Leftrightarrow x - 2 \ \textgreater \ 0 \\ \\ \Leftrightarrow \boxed{x \ \textgreater \ 2} [/tex]
Therefore, f is concave up at [tex](2, \infty)[/tex].
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
[tex]f''(x) = 30x - 60[/tex]
Using the similar computation in (c),
[tex]f''(x) \ \textless \ 0 \\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 \\ \\ \Leftrightarrow x - 2 \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}[/tex]
Therefore, f is concave down at [tex](-\infty, 2)[/tex].
(a) f is increasing at [tex](-\infty,-2) \cup (2,\infty)[/tex].
(b) f is decreasing at [tex](-2,2)[/tex].
(c) f is concave up at [tex](2, \infty)[/tex]
(d) f is concave down at [tex](-\infty, 2)[/tex]
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
[tex]f'(x) = 15x^2 - 60 [/tex]
So,
[tex] f'(x) \ \textgreater \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}[/tex]
The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:
-->> x < -2
-->> -2 < x < 2
--->> x > 2
If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.
If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.
So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since
[tex]f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0[/tex]
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at [tex](-\infty,-2) \cup (2,\infty)[/tex].
(b) f is decreasing only when the derivative of f is negative. Since
[tex]f'(x) = 15x^2 - 60 [/tex]
Using the similar computation in (a),
[tex]f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}[/tex]
Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.
Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)
(c) f is concave up if and only if the second derivative of f is positive. Note that
[tex]f''(x) = 30x - 60[/tex]
Since,
[tex]f''(x) \ \textgreater \ 0 \\ \\ \Leftrightarrow 30x - 60 \ \textgreater \ 0 \\ \\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0 \\ \\ \Leftrightarrow x - 2 \ \textgreater \ 0 \\ \\ \Leftrightarrow \boxed{x \ \textgreater \ 2} [/tex]
Therefore, f is concave up at [tex](2, \infty)[/tex].
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
[tex]f''(x) = 30x - 60[/tex]
Using the similar computation in (c),
[tex]f''(x) \ \textless \ 0 \\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 \\ \\ \Leftrightarrow x - 2 \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}[/tex]
Therefore, f is concave down at [tex](-\infty, 2)[/tex].
