let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(3)=7 )…

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(3)=7 ) and ( 2 leq f^{prime}(x) leq 5 ) for all ( -infty < x < infty ). what is the range of possible values for ( f(8) ) and ( f(0) )? fill in the blanks below:\n1. ( quad leq f(8) leq )\n2. ( quad leq f(0) leq )

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(3)=7 ) and ( 2 leq f^{prime}(x) leq 5 ) for all ( -infty < x < infty ). what is the range of possible values for ( f(8) ) and ( f(0) )? fill in the blanks below:\n1. ( quad leq f(8) leq )\n2. ( quad leq f(0) leq )

Answer

Explanation:

Step1: Apply the Mean - Value Theorem

By the Mean - Value Theorem, (f(b)-f(a)=f^{\prime}(c)(b - a)) for some (c\in(a,b)).

For (f(8)) and (a = 3), (b=8), then (f(8)-f(3)=f^{\prime}(c)(8 - 3)=5f^{\prime}(c)), since (f(3) = 7), we have (f(8)=7 + 5f^{\prime}(c)).

Step2: Find the range of (f(8))

Given (2\leq f^{\prime}(x)\leq5). When (f^{\prime}(c)=2), (f(8)=7+5\times2=17). When (f^{\prime}(c)=5), (f(8)=7 + 5\times5=32). So (17\leq f(8)\leq32).

Step3: For (f(0)) and (a = 3), (b = 0)

(f(0)-f(3)=f^{\prime}(c)(0 - 3)=- 3f^{\prime}(c)), since (f(3)=7), then (f(0)=7-3f^{\prime}(c)).

Step4: Find the range of (f(0))

When (f^{\prime}(c)=2), (f(0)=7-3\times2=1). When (f^{\prime}(c)=5), (f(0)=7-3\times5=-8). So (-8\leq f(0)\leq1).

Answer:

  1. (17\leq f(8)\leq32)
  2. (-8\leq f(0)\leq1)