A paint can has a radius of 4 inches and a height of 15 inches. What is the volume of the paint can? Round to the nearest tenth.
2 answers:
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Answers:
(a) f is increasing at
.
(b) f is decreasing at
.
(c) f is concave up at
(d) f is concave down at
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
So,
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
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at.
(b) f is decreasing only when the derivative of f is negative. Since
Using the similar computation in (a),
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
Since,
Therefore, f is concave up at.
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
Using the similar computation in (c),
Therefore, f is concave down at.
(a) f is increasing at
(b) f is decreasing at
(c) f is concave up at
(d) f is concave down at
Explanations:
(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that
So,
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
Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at
(b) f is decreasing only when the derivative of f is negative. Since
Using the similar computation in (a),
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
Since,
Therefore, f is concave up at
(d) Note that f is concave down if and only if the second derivative of f is negative. Since,
Using the similar computation in (c),
Therefore, f is concave down at