- Over the interval [-3, 1], the local minimum is 0, since [tex]f(-2) < f(x),\forall \,x \neq -2, -3 \le x \le -1[/tex]. Right choice: D.
- Over the interval [-1, 0], the local maximum is 4.39, since [tex]f(-0.8) > f(x),\forall \,x \neq -0.8, -1 \le x \le 0[/tex]. Right choice: D.
- Over the interval [0, 3], the local minimum is -32, since [tex]f(2) < f(x), \forall \,x \ne 2 , 0 \le x\le 3[/tex]. Right choice: B.
Given a function that is continuous for an interval [tex][a, b][/tex], [tex]a \neq b[/tex], then there is a local minimum and local maximum if and only if:
[tex]f(c) < f(x), \forall \,x \neq c[/tex], [tex]a \leq c \leq b[/tex] (1)
[tex]f(d) > f(x), \forall x \neq d, a \le d \le b[/tex] (2)
Based on these facts, we proceed to solve each question:
- Over the interval [-3, 1], the local minimum is 0, since [tex]f(-2) < f(x),\forall \,x \neq -2, -3 \le x \le -1[/tex]. Right choice: D.
- Over the interval [-1, 0], the local maximum is 4.39, since [tex]f(-0.8) > f(x),\forall \,x \neq -0.8, -1 \le x \le 0[/tex]. Right choice: D.
- Over the interval [0, 3], the local minimum is -32, since [tex]f(2) < f(x), \forall \,x \ne 2 , 0 \le x\le 3[/tex]. Right choice: B.
We kindly invite to check this question on maxima and minima: https://brainly.com/question/2292974