Question

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreasing. (-2,2) find the intervals on which f is concave up. (-inf,-2)u(2,inf) find the intervals on which f is concave down.

Answer 1

Answers:

(a) f is increasing at
`(-\infty,-2) \cup (2,\infty)`.

(b) f is decreasing at
`(-2,2)`.

(c) f is concave up at
`(2, \infty)`

(d) f is concave down at
`(-\infty, 2)`

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that


`f'(x) = 15x^2 - 60`

So,


`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)}`

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


`f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0`

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at
`(-\infty,-2) \cup (2,\infty)`.

(b) f is decreasing only when the derivative of f is negative. Since


`f'(x) = 15x^2 - 60`

Using the similar computation in (a),


`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)}`

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


`f''(x) = 30x - 60`

Since,


`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}`

Therefore, f is concave up at
`(2, \infty)`.

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,


`f''(x) = 30x - 60`

Using the similar computation in (c),


`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}`

Therefore, f is concave down at
`(-\infty, 2)`.