Let f be a twice differentiable function, and let f(-3)=-4, f^(')(-3)=0, and f^('')(-3)=1. What occurs in the graph of f at the point (-3,-4) ? Choose 1 answer: (A) (-3,-4) is a minimum point. (B) (-3,-4) is a maximum point. (C) There's not enough information to tell.

Question

Let  f be a twice differentiable function, and let  f(-3)=-4,  f^(')(-3)=0, and  f^('')(-3)=1. What occurs in the graph of  f at the point  (-3,-4) ? Choose 1 answer: (A)  (-3,-4) is a minimum point. (B)  (-3,-4) is a maximum point. (C) There's not enough information to tell.

Bytelearn · Accepted Answer

Given Information: Given that f is a twice differentiable function, we have information about the function and its first two derivatives at the point x = -3. We know that f(-3) = -4, f'(-3) = 0, and f''(-3) = 1.Point on Graph: The value f(-3) = -4 tells us that the point (-3, -4) lies on the graph of the function f. However, this information alone does not tell us about the nature of the point on the graph.Tangent Line: The derivative f'(-3) = 0 indicates that the slope of the tangent to the graph of f at x = -3 is zero. This means that the graph of f has a horizontal tangent line at the point (-3, -4). This could be indicative of a local maximum, a local minimum, or a point of inflection.Concavity Analysis: The second derivative f''(-3) = 1 tells us about the concavity of the graph at x = -3. Since f''(-3) is positive, the graph of f is concave up at this point. This suggests that the point (-3, -4) is a local minimum.Conclusion: Combining the information from the first and second derivatives, we can conclude that the point (-3, -4) is a local minimum on the graph of f because the first derivative is zero (indicating a horizontal tangent) and the second derivative is positive (indicating concave up).

Full solution

More problems from Nth roots

Related topics