If f(2) = 12 and f '(x) ≥ 3 for 2 ≤ x ≤ 5, how small can f(5) possibly be?
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f(2) = 12, and f'(x) >= 3 for 2 <= x <= 5, how small can f(5) possibly be.
Keep in mind that f'(x) represents the slope at a given point.
Since the slope is at the very least 3, we know that slope represents how fast a function increases.
Since the smallest slope is 3, let's say the slope is 3.
If we say the slope is 3, we now want to find the value at x = 5 with whatever line represents the one that crosses (2, 12) with a slope of 3. Find the equation of the line with slope m = 3 through (2, 12).
(y2 - y1)/(x2 - x1) = m
Let m = 3, (x1, y1) = (2, 12), and (x2, y2) = (x, y).
(y - 12)/(x - 2) = 3
y - 12 = 3(x - 2)
y - 12 = 3x - 6
y = 3x + 6
So we can calculate f(5) by letting x = 5.
y = 3(5) + 6
y = 15 + 6
y = 21
The smallest value y can be is 21, as we use the function f(x) = 3x + 6 being the most extreme example (since f'(x) is exactly 3 everywhere, including 2 <= x <= 5).
Keep in mind that f'(x) represents the slope at a given point.
Since the slope is at the very least 3, we know that slope represents how fast a function increases.
Since the smallest slope is 3, let's say the slope is 3.
If we say the slope is 3, we now want to find the value at x = 5 with whatever line represents the one that crosses (2, 12) with a slope of 3. Find the equation of the line with slope m = 3 through (2, 12).
(y2 - y1)/(x2 - x1) = m
Let m = 3, (x1, y1) = (2, 12), and (x2, y2) = (x, y).
(y - 12)/(x - 2) = 3
y - 12 = 3(x - 2)
y - 12 = 3x - 6
y = 3x + 6
So we can calculate f(5) by letting x = 5.
y = 3(5) + 6
y = 15 + 6
y = 21
The smallest value y can be is 21, as we use the function f(x) = 3x + 6 being the most extreme example (since f'(x) is exactly 3 everywhere, including 2 <= x <= 5).