Question

let ( f(x) ) be a continuous and differentiable function for all ( -infty < x < infty ).
suppose ( f(1)=5 ) and ( f^{prime}(x) geq 3 ) for all ( -infty < x < infty ).

1. what is the smallest possible value for ( f(7) )?

answer:

1. what is the largest possible value for ( f(-2) )?

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. \(-4\)