let /be a continuous function on the closed interval -3,6. if $f(-3)=-1$ and $f(6)=3$, then the intermediate…

let /be a continuous function on the closed interval -3,6. if $f(-3)=-1$ and $f(6)=3$, then the intermediate value theorem guarantees that\na $f(0)=0$\nb $f(c)=\\frac{4}{9}$ for at least one c between -3 and 6\nc $-1\\leq f(x)\\leq 3$ for all x between -3 and 6\nd $f(c)=1$ for at least one c between -3 and 6\ne $f(c)=0$ for at least one c between -1 and 3

let /be a continuous function on the closed interval -3,6. if $f(-3)=-1$ and $f(6)=3$, then the intermediate value theorem guarantees that\na $f(0)=0$\nb $f(c)=\\frac{4}{9}$ for at least one c between -3 and 6\nc $-1\\leq f(x)\\leq 3$ for all x between -3 and 6\nd $f(c)=1$ for at least one c between -3 and 6\ne $f(c)=0$ for at least one c between -1 and 3

Answer

Explanation:

Step1: Recall the Intermediate Value Theorem

The Intermediate Value Theorem states that if (y = f(x)) is continuous on a closed interval ([a,b]), and (k) is a number between (f(a)) and (f(b)) ((f(a)\neq f(b))), then there exists at least one number (c\in(a,b)) such that (f(c)=k). Here, (a=-3), (b = 6), (f(-3)=-1), and (f(6)=3). The value (k = 1) is between (-1) and (3) ((-1<1<3)).

Step2: Analyze each option

  • Option A: The Intermediate Value Theorem does not guarantee that (f(0)=0). Just because the function is continuous on ([-3,6]) and (f(-3)=-1), (f(6) = 3), there is no basis to say (f(0)=0).
  • Option B: (f^{\prime}(c)=\frac{4}{9}) is related to the Mean Value Theorem ((f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}=\frac{3-(-1)}{6-(-3)}=\frac{4}{9}) when (f) is differentiable on ((-3,6)) and continuous on ([-3,6])), not the Intermediate Value Theorem.
  • Option C: The Extreme Value Theorem (if (f) is continuous on ([a,b]), then (f) has a maximum (M) and a minimum (m) on ([a,b]) such that (m\leq f(x)\leq M) for all (x\in[a,b])) is needed to guarantee bounds for all (x) in the interval, not the Intermediate Value Theorem.
  • Option D: Since (f) is continuous on ([-3,6]), (f(-3)=-1), (f(6)=3), and (1) is between (-1) and (3), by the Intermediate Value Theorem, there exists at least one (c\in(-3,6)) such that (f(c)=1).
  • Option E: The interval ([-1,3]) is not the interval ([-3,6]) on which we know the function is continuous. The Intermediate Value Theorem is applied on the interval where the function is given to be continuous (([-3,6]) in the problem statement).

Answer:

D. (f(c)=1) for at least one (c) between (-3) and (6)