Determine the interval(s) at which f(x) is concave up given that f′′(x)=−x2+x+6.
Critical points are: -2 & 3
To be concave up the 2nd derivative must be positive, so
-x^2 + x + 6 > 0
x^2 - x - 6 < 0
(x-3)(x+2) < 0
looking at y = x^2 - x - 6 , we see from our solution above that it has x-intercepts of 3 and -2, thus it is above ( greater than zero) for
x < -2 or x > 3
To determine the interval(s) at which a function f(x) is concave up, we need to analyze the second derivative of the function. The second derivative tells us the rate of change of the first derivative and provides information about the concavity of the function.
Given that f′′(x) = -x^2 + x + 6, we can use this information to find the intervals where f(x) is concave up.
Step 1: Find the critical points.
To determine the intervals, we need to find the critical points of the function. The critical points are the values of x where the first derivative is equal to zero or undefined.
In this case, the critical points are -2 and 3 (found based on the given second derivative), where f'(x) = 0.
Step 2: Analyze the second derivative.
We want to determine the intervals where f(x) is concave up, meaning where the second derivative is positive.
The second derivative f''(x) = -x^2 + x + 6 is a quadratic equation. For a quadratic to be positive, the value of the quadratic expression must be greater than zero.
To find the intervals where f(x) is concave up, we need to solve the inequality: -x^2 + x + 6 > 0.
Step 3: Solve the inequality.
To solve the inequality, we need to find the values of x that make the expression -x^2 + x + 6 greater than zero.
To do this, we can factor the quadratic expression or use the quadratic formula. However, in this case, it's easier to determine the intervals by graphing or using the sign chart method.
Since the quadratic equation has a negative coefficient on the x^2 term (-x^2), the graph of the function will open downwards. That means the quadratic function will be positive outside of the interval between the critical points.
By graphing or using the sign chart, we can determine that the quadratic function is positive for all x values outside of the interval (-2, 3).
Step 4: Determine the intervals.
Based on our analysis, we found that the quadratic function is positive for all x values outside of the interval (-2, 3). This means that f(x) is concave up for all x values less than -2 and greater than 3.
Therefore, the interval(s) at which f(x) is concave up is (-∞, -2) U (3, +∞).