let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(1)=5 )…

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(1)=5 ) and ( f^{prime}(x) geq 3 ) for all ( -infty < x < infty ).\n1. what is the smallest possible value for ( f(7) )?\nanswer:\n2. what is the largest possible value for ( f(-2) )?\nanswer:

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).\nsuppose ( f(1)=5 ) and ( f^{prime}(x) geq 3 ) for all ( -infty < x < infty ).\n1. what is the smallest possible value for ( f(7) )?\nanswer:\n2. what is the largest possible value for ( f(-2) )?\nanswer:

Answer

Explanation:

Step1: Apply the Mean Value Theorem

The Mean Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and differentiable on the open interval ((a,b)), then (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}) for some (c\in(a,b)).

For (f(7)):

Let (a = 1), (b=7). Then (\frac{f(7)-f(1)}{7 - 1}=f^{\prime}(c)), where (c\in(1,7)). Since (f^{\prime}(x)\geq3) for all (x), we have (\frac{f(7)-5}{6}\geq3).

For (f(- 2)):

Let (a=-2), (b = 1). Then (\frac{f(1)-f(-2)}{1-(-2)}=f^{\prime}(c)), where (c\in(-2,1)). Since (f^{\prime}(x)\geq3) for all (x), we have (\frac{5 - f(-2)}{3}\geq3).

Step2: Solve the inequalities

For (f(7)):

Multiply both sides of (\frac{f(7)-5}{6}\geq3) by (6): (f(7)-5\geq18). Then add (5) to both sides: (f(7)\geq18 + 5=23).

For (f(-2)):

Multiply both sides of (\frac{5 - f(-2)}{3}\geq3) by (3): (5 - f(-2)\geq9). Subtract (5) from both sides: (-f(-2)\geq9 - 5 = 4). Multiply both sides by (- 1) (and reverse the inequality sign): (f(-2)\leq1).

Answer:

  1. (23)
  2. (-4)