Question

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).
suppose ( 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:

1. ( quad leq f(8) leq )
2. ( quad leq f(0) leq )

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. \(-8\leq f(0)\leq1\)