Suppose that f is a continuous function that is differentiable everywhere. The graph of the derivative of f crosses the x-axis three times. What is the minimum number of intervals of (−∞,∞) where f has tangent lines with positive slope?
Since the graph of the derivative of f crosses the x-axis three times, this means that the derivative of f changes sign three times. Therefore, f(x) must have either a local maximum or a local minimum at each point where its derivative changes sign.
Let's consider the points where the derivative changes sign. At each of these points, f(x) must either have a local maximum or a local minimum. However, to have a tangent line with positive slope at a point on the graph of f, f(x) must have a local minimum at that point.
Therefore, for each sign change of the derivative of f, there must be at least one interval where f has a tangent line with positive slope. Since the derivative of f changes sign three times, there must be at least three intervals where f has a tangent line with positive slope.
Therefore, the minimum number of intervals of (−∞,∞) where f has tangent lines with positive slope is three.
Suppose that f is a continuous function that is differentiable everywhere. The graph of the derivative of f crosses the x-axis three times. What is the minimum number of intervals of (−∞,∞) where f has tangent lines with positive slope? (1 point) Responses
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The minimum number of intervals of (−∞,∞) where f has tangent lines with positive slope is 2.
Let f(x)={9x+7, x≤−1ax2+bx, x>−1 . Find a and b such that f is continuous and differentiable on (−∞,∞) .
For f(x) to be continuous at x = -1, we need the left and right-hand limits to be equal.
Therefore, we have:
lim(x→-1-)(9x + 7) = lim(x→-1+)(ax^2 + bx)
Substituting x = -1 into 9x + 7, we get:
-9 + 7 = a(-1)^2 + b(-1)
-2 = a - b
Next, to ensure f(x) is differentiable at x = -1, we need the left and right-hand derivatives to be equal.
The derivative of 9x + 7 is 9.
The derivative of ax^2 + bx is 2ax + b.
Therefore, we have:
9 = 2a(0) + b
9 = b
Now we can substitute b = 9 into the equation -2 = a - b:
-2 = a - 9
a = 7
Therefore, a = 7 and b = 9 for f to be continuous and differentiable on (-∞,∞).
To determine the minimum number of intervals where f has tangent lines with a positive slope, we need to analyze the behavior of the derivative of f.
Given that the graph of the derivative of f crosses the x-axis three times, we can conclude that there are at least three critical points (points where the derivative equals zero or is undefined) on f.
Let's consider the possible scenarios:
1. If the derivative changes from positive to negative or negative to positive at each of the critical points, then f would have tangent lines with a positive slope on both sides of each critical point. In this case, there would be three intervals where f has tangent lines with positive slope.
2. If the derivative changes from positive to negative at each critical point, or negative to positive at each critical point, then f would have tangent lines with a positive slope on only one side of each critical point. In this case, there would be two intervals where f has tangent lines with positive slope.
3. If the derivative remains positive or negative throughout all the critical points, then f would have tangent lines with a positive slope on either the left or the right side of all critical points. In this case, there would be one interval where f has tangent lines with positive slope.
Therefore, the minimum number of intervals of (-∞,∞) where f has tangent lines with a positive slope is one.
To determine the minimum number of intervals where f has tangent lines with positive slope, we need to analyze the relationship between the original function f and its derivative.
1. The graph of the derivative, f'(x), crossing the x-axis three times indicates that there are three roots or zeros of f'(x).
2. Each root of f'(x) represents a critical point on the graph of f(x). At these points, the slope of the tangent line will be zero.
3. Considering only the intervals between these critical points, we need to examine the sign of f'(x) within each interval to determine the slope of the tangent lines.
4. By the Intermediate Value Theorem, if f'(x) changes sign from negative to positive within an interval, it implies that there is at least one point within that interval where the tangent line has a positive slope.
Therefore, the minimum number of intervals where f has tangent lines with positive slope is equal to the number of changes in sign of f'(x).
Thus, the answer is three, since we know that f'(x) crosses the x-axis three times, indicating three sign changes.