Let g be a twice differentiable function, and let g(−6)=−1, g′(−6)=0, and g′′(−6)=−3.What occurs in the graph of g at the point (−6,−1) ?Choose 1 answer:(A) (−6,−1) is a minimum point.(B) (−6,−1) is a maximum point.(C) There's not enough information to tell.
Q. Let g be a twice differentiable function, and let g(−6)=−1, g′(−6)=0, and g′′(−6)=−3.What occurs in the graph of g at the point (−6,−1) ?Choose 1 answer:(A) (−6,−1) is a minimum point.(B) (−6,−1) is a maximum point.(C) There's not enough information to tell.
Analyze Given Information: First, we analyze the given information about the function g at the point x=−6. We are given that g(−6)=−1, which means the point (−6,−1) lies on the graph of g.
Consider First Derivative: Next, we consider the first derivative g′(−6)=0. This indicates that the slope of the tangent to the graph of g at x=−6 is zero, which means the graph has a horizontal tangent line at this point. This could be indicative of a local maximum, local minimum, or a saddle point (inflection point).
Look at Second Derivative: We then look at the second derivative g′′(−6)=−3. Since the second derivative is negative, it tells us that the graph of g is concave down at x=−6. This concavity, combined with the horizontal tangent line, means that the point (−6,−1) is a local maximum.
Conclude Maximum Point: Given the information about the first and second derivatives at x=−6, we can conclude that (−6,−1) is a maximum point on the graph of g.