If a number is divisible by both 2 and 3 then we can say the number is divisible by A. 5. B. 2. C. 4. D. 6.
2 answers:
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Slope intercept form: y = mx + b
mx = slope
b = y-intercept
We know the y intercept is 0, so nothing will be written there.
To find the slope of this line, we can use the slope formula.
We'll use the points (1, 0) and (3, 1) to find the slope.
Now we can just plug these values into the equation to find the slope.
1 - 0 / 3 - 1
1 / 2
The slope of the line is 1/2, or 0.5.
The slope-intercept form of this line can be written as:
y = 0.5x
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