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?
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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 (-∞,∞).
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 (-∞,∞).