Classify the stationary point of the following graph by evaluating the coordinates 1 1 on either side of the stationary point: � ( � ) = 1.5 � 2 − 6 � + 4 f(x)=1.5x 2 −6x+4 This function has a stationary point at ( 2 , − 2 ) (2,−2) . One minimum turning point. One rising point of inflection. One maximum turning point. One falling point of inflection.
Question
Classify the stationary point of the following graph by evaluating the coordinates 1 1 on either side of the stationary point:
� ( � )
1.5 � 2 − 6 � + 4 f(x)=1.5x 2 −6x+4
This function has a stationary point at ( 2 , − 2 ) (2,−2) .
One minimum turning point.
One rising point of inflection.
One maximum turning point.
One falling point of inflection.
Solution
To classify the stationary point of the function f(x)=1.5x^2 -6x+4, we first need to find the derivative of the function. The derivative of the function is f'(x)=3x-6.
Setting the derivative equal to zero gives us the x-coordinate of the stationary point: 3x-6=0, so x=2. Substituting x=2 into the original function gives us the y-coordinate of the stationary point: f(2)=1.5(2)^2 -6(2)+4 = -2. So the stationary point is indeed at (2,-2).
To classify the stationary point, we need to look at the second derivative of the function. The second derivative is f''(x)=3. Since the second derivative is positive, the stationary point at (2,-2) is a minimum turning point.
So the correct answer is: One minimum turning point.
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