Suppose that 2 ≤ f '(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of f(9) − f(4)?

Answer 1

To find the minimum and maximum possible values of \(f(9) - f(4)\) given that \(2 \leq f'(x) \leq 5\) for all values of \(x\), we can use the Mean Value Theorem.

The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

In this case, let \(a = 4\) and \(b = 9\). Therefore:

\[ f'(c) = \frac{f(9) - f(4)}{9 - 4} \]

Rearranging to solve for \(f(9) - f(4)\), we get:

\[ f(9) - f(4) = 5 \cdot 5 \]

So, the minimum value of \(f(9) - f(4)\) occurs when \(f'(c) = 2\) and the maximum value occurs when \(f'(c) = 5\).

Therefore, the minimum value is \(5 \cdot 2 = 10\) and the maximum value is \(5 \cdot 5 = 25\). Hence, \(10 \leq f(9) - f(4) \leq 25\).

Answer 2

2<f'(x)<5

integrate, ignore the constant of integration, as it will cancel out in the end

2x < F(x) < 5x

2(9)=18 < F(9)< 5(9)=45

2(4)=8< F(4)<5(4)=20

subtract

18-8=10< F(9)-F(4)< 45-20=25

10=minimum F(9)-F(4)

25 = maximum F(9)-F(4)

underline all the less than signs above