The graph of f′′(x)f''(x)f′′(x) is shown below. Use the graph to determine the intervals for which f(x)f\left(x\right)f(x)is concave up or concave down and the location of any inflection points.

Concave up: ( − ∞ , 0 ) \left(-\infty,0\right) ( − ∞ , 0 ) , ( 2 , ∞ ) \left(2,\infty\right) ( 2 , ∞ ) ; Concave down: ( 0 , 2 ) \left(0,2\right) ( 0 , 2 ) ; Inflection points: x = 0 x=0 x = 0 , x = 2 x=2 x = 2

The graph of f ′ ′ ( x ) f''(x) f ′′ ( x ) is shown below. Use the graph to determine the intervals for which f ( x ) f\left(x\right) f ( x ) is concave up or concave down and the location of any inflection points.

Here we are shown the graph of f double prime of x. That's the second derivative of f of x. Now we want to use this graph to determine the intervals for which f of x is concave up or concave down and the location of any inflection points. Now when we're given the graph of a second derivative of a function, in order to determine where that original function, f of x, is concave up or concave down, we are just looking for whether our second derivative is positive or negative, above or below the x axis. So my function will be concave up where that second derivative is positive and concave down where that second derivative is negative. Our inflection points will then be located wherever the sign changes. So this is realistically where f double prime crosses the x axis, transitions from being positive to negative or from negative to positive. So let's go ahead and take a look at our graph here remembering that this is the graph of f double prime and not just the graph of f. Now we wanna look here for where it's positive or negative. Now it starts off being positive, so we know that it is con that our original function is concave up from negative infinity until this graph becomes negative, which we see happens here when we cross over into the into below our x axis. So from negative infinity until x equals 0, my function is concave up. Then we can see that our derivative here is negative from 0 until we reach 2. That's when it stops being below the x axis. So it's concave down from x equals 0 to x equals 2. Then it becomes positive again over here from 2 on to infinity. So we have all of our intervals here. Let's go ahead and identify our inflection points. Remember, this is going to be where f double prime crosses the x axis, so where it changes from being positive to negative or negative to positive. We know that these will be the points in between all of the intervals that we identified. We can see that crosses the x axis here at x equals 0 and then again here at x equals 2. Now remember that we can't tell the entire ordered pair of our inflection point. I can't say with certainty that my inflection point is located at 0 or 20 because this is the graph of the second derivative. It's not the graph of the function itself. I do know the location of these in terms of my x value but not the actual function value there. So now we have our intervals of concavity and our inflection points based here on our second derivative. Let me know if you have any questions and let's keep going.

The graph of f ′ ′ ( x ) f''(x) f ′′ ( x ) is shown below. Use the graph to determine the intervals for which f ( x ) f\left(x\right) f ( x ) is concave up or concave down and the location of any inflection points.

Concave up: ( − ∞ , 0 ) \left(-\infty,0\right) ( − ∞ , 0 ) , ( 2 , ∞ ) \left(2,\infty\right) ( 2 , ∞ ) ; Concave down: ( 0 , 2 ) \left(0,2\right) ( 0 , 2 ) ; Inflection points: x = 0 x=0 x = 0 , x = 2 x=2 x = 2

Concave down: ( − 1 , 0 ) \left(-1,0\right) ( − 1 , 0 ) , ( 2 , ∞ ) \left(2,\infty\right) ( 2 , ∞ ) ; Concave up: ( 0 , 2 ) \left(0,2\right) ( 0 , 2 ) ; Inflection points: x = 0 x=0 x = 0 , x = 2 x=2 x = 2

Concave down: ( − 1 , 0 ) \left(-1,0\right) ( − 1 , 0 ) , ( 2 , ∞ ) \left(2,\infty\right) ( 2 , ∞ ) ; Concave up: ( 0 , 2 ) \left(0,2\right) ( 0 , 2 ) ; Inflection points: x = − 1 x=-1 x = − 1 , x = 0 x=0 x = 0 , x = 2 x=2 x = 2

Concave up: ( − ∞ , − 1 ) \left(-\infty,-1\right) ( − ∞ , − 1 ) , ( 0 , ∞ ) \left(0,\infty\right) ( 0 , ∞ ) ; Concave down: ( − 1 , 0 ) \left(-1,0\right) ( − 1 , 0 ) ; Inflection points: x = − 1 x=-1 x = − 1 , x = 0 x=0 x = 0

5. Graphical Applications of Derivatives

Concavity

Calculus

5. Graphical Applications of Derivatives

Concavity

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Multiple Choice

The graph of $f''(x)$ is shown below. Use the graph to determine the intervals for which $f\left(x\right)$ is concave up or concave down and the location of any inflection points.

Concave down: $\left(-1,0\right)$ , $\left(2,\infty\right)$ ; Concave up: $\left(0,2\right)$ ; Inflection points: $x=0$ , $x=2$

Concave down: $\left(-1,0\right)$ , $\left(2,\infty\right)$ ; Concave up: $\left(0,2\right)$ ; Inflection points: $x=-1$ , $x=0$ , $x=2$

Concave up: $\left(-\infty,0\right)$ , $\left(2,\infty\right)$ ; Concave down: $\left(0,2\right)$ ; Inflection points: $x=0$ , $x=2$

Concave up: $\left(-\infty,-1\right)$ , $\left(0,\infty\right)$ ; Concave down: $\left(-1,0\right)$ ; Inflection points: $x=-1$ , $x=0$

Verified step by step guidance

To determine the concavity of f(x), we need to analyze the graph of f''(x). The graph provided is a parabola opening upwards, which represents f''(x).

The function f(x) is concave up where f''(x) > 0. From the graph, f''(x) is positive for x < 0 and x > 2.

The function f(x) is concave down where f''(x) < 0. From the graph, f''(x) is negative for 0 < x < 2.

Inflection points occur where f''(x) changes sign. From the graph, this happens at x = 0 and x = 2.

Thus, the intervals of concavity and inflection points are: Concave up on (-∞, 0) and (2, ∞); Concave down on (0, 2); Inflection points at x = 0 and x = 2.

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