Does ƒ(x) = (x⁶/2) + (5x⁴/4) - 15x² have any inflection points? If so, identify them.
Ch. 4 - Applications of the Derivative
4장, 문제 20
Use ƒ' and ƒ" to complete parts (a) and (b).
a. Find the intervals on which f is increasing and the intervals on which it is decreasing.
b. Find the intervals on which f is concave up and the intervals on which it is concave down.
ƒ(x) = x√(x +9)
검증된 단계별 안내1
Step 1: Find the first derivative ƒ'(x) of the function ƒ(x) = x√(x + 9). Use the product rule and chain rule to differentiate. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is u'(x)v(x) + u(x)v'(x). Here, let u(x) = x and v(x) = √(x + 9).
Step 2: Simplify the expression for ƒ'(x) and set it equal to zero to find the critical points. These critical points will help determine where the function is increasing or decreasing. Solve the equation ƒ'(x) = 0 for x.
Step 3: Use the critical points to test intervals on the number line. Choose test points in each interval and substitute them into ƒ'(x) to determine the sign of the derivative. If ƒ'(x) > 0, the function is increasing on that interval; if ƒ'(x) < 0, the function is decreasing.
Step 4: Find the second derivative ƒ''(x) to analyze the concavity of the function. Differentiate ƒ'(x) to obtain ƒ''(x). This will involve applying the product rule and chain rule again.
Step 5: Determine the intervals of concavity by setting ƒ''(x) equal to zero and solving for x to find possible inflection points. Test intervals around these points using test values to determine the sign of ƒ''(x). If ƒ''(x) > 0, the function is concave up on that interval; if ƒ''(x) < 0, the function is concave down.

비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
10m주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
First Derivative Test
The first derivative of a function, denoted as ƒ', provides information about the function's increasing and decreasing behavior. If ƒ' > 0 on an interval, the function is increasing; if ƒ' < 0, it is decreasing. By finding critical points where ƒ' = 0 or is undefined, we can determine the intervals of increase and decrease.
추천 영상:
The First Derivative Test: Finding Local Extrema
Second Derivative Test
The second derivative of a function, denoted as ƒ'', indicates the concavity of the function. If ƒ'' > 0, the function is concave up, suggesting that the slope of the tangent line is increasing. Conversely, if ƒ'' < 0, the function is concave down, indicating that the slope is decreasing. Analyzing points where ƒ'' = 0 helps identify inflection points.
추천 영상:
The Second Derivative Test: Finding Local Extrema
Critical Points
Critical points occur where the first derivative is zero or undefined, and they are essential for determining intervals of increase and decrease. These points can indicate local maxima or minima. Additionally, critical points are also relevant for the second derivative test, as they may correspond to changes in concavity, helping to identify inflection points.
추천 영상:
Critical Points
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