question 11 (1 point)\nsuppose a function f(x) is differentiable everywhere and has a local minimum at x =…

question 11 (1 point)\nsuppose a function f(x) is differentiable everywhere and has a local minimum at x = c.\nif f(x) < 0 when x < c, and f(x) > 0 when x > c, then by the global interval method we\nknow x = c is\na local minimum\nan absolute maximum\nan absolute minimum\na local maximum\nview hint for question 11\nlevel 2: advanced problems

question 11 (1 point)\nsuppose a function f(x) is differentiable everywhere and has a local minimum at x = c.\nif f(x) < 0 when x < c, and f(x) > 0 when x > c, then by the global interval method we\nknow x = c is\na local minimum\nan absolute maximum\nan absolute minimum\na local maximum\nview hint for question 11\nlevel 2: advanced problems

Answer

Brief Explanations:

According to the first - derivative test for local extrema: If a function (y = f(x)) is differentiable at (x = c), and (f^{\prime}(x)) changes sign from negative to positive as (x) increases through (c) (i.e., (f^{\prime}(x)<0) for (x < c) and (f^{\prime}(x)>0) for (x>c)), then (x = c) is a local minimum of the function (y = f(x)). The global interval method is related to the first - derivative test. An absolute maximum/minimum is the largest/smallest value of the function over its entire domain (not just in a local neighborhood), and we have no information about the function's behavior over its entire domain. A local maximum would require (f^{\prime}(x)>0) for (x < c) and (f^{\prime}(x)<0) for (x>c).

Answer:

a local minimum