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Math  /  Calculus

Question23. If f(1)=10f(1)=10 and f′(x)⩾2f^{\prime}(x) \geqslant 2 for 1⩽x⩽41 \leqslant x \leqslant 4, how small can f(4)f(4) possibly be?
24. Suppose that 3⩽f′(x)⩽53 \leqslant f^{\prime}(x) \leqslant 5 for all values of xx. Show that 18⩽f(8)−f(2)⩽3018 \leqslant f(8)-f(2) \leqslant 30.
25. Does there exist a function ff such that f(0)=−1,f(2)=4f(0)=-1, f(2)=4, and f′(x)⩽2f^{\prime}(x) \leqslant 2 for all xx ?

Studdy Solution

STEP 1

1. For problem 23, we know f(1)=10 f(1) = 10 and f′(x)≥2 f'(x) \geq 2 for 1≤x≤4 1 \leq x \leq 4 .
2. For problem 24, 3≤f′(x)≤5 3 \leq f'(x) \leq 5 for all x x .
3. For problem 25, we need to determine if a function f f exists with f(0)=−1 f(0) = -1 , f(2)=4 f(2) = 4 , and f′(x)≤2 f'(x) \leq 2 for all x x .

STEP 2

1. Solve problem 23 by using the Mean Value Theorem.
2. Solve problem 24 by applying the Mean Value Theorem and bounding the change in f(x) f(x) .
3. Analyze problem 25 by considering the implications of the derivative constraint.

STEP 3

For problem 23, apply the Mean Value Theorem which states that there exists some c c in (1,4) (1, 4) such that:
f′(c)=f(4)−f(1)4−1 f'(c) = \frac{f(4) - f(1)}{4 - 1}
Since f′(x)≥2 f'(x) \geq 2 for 1≤x≤4 1 \leq x \leq 4 , we have:
f(4)−103≥2 \frac{f(4) - 10}{3} \geq 2

STEP 4

Solve the inequality:
f(4)−10≥6 f(4) - 10 \geq 6
f(4)≥16 f(4) \geq 16
Thus, the smallest possible value for f(4) f(4) is:
16 \boxed{16}

STEP 5

For problem 24, apply the Mean Value Theorem which states that there exists some c c in (2,8) (2, 8) such that:
f′(c)=f(8)−f(2)8−2 f'(c) = \frac{f(8) - f(2)}{8 - 2}
Given 3≤f′(x)≤5 3 \leq f'(x) \leq 5 , we have:
3≤f(8)−f(2)6≤5 3 \leq \frac{f(8) - f(2)}{6} \leq 5

STEP 6

Multiply the entire inequality by 6:
18≤f(8)−f(2)≤30 18 \leq f(8) - f(2) \leq 30
Thus, the range for f(8)−f(2) f(8) - f(2) is:
18≤f(8)−f(2)≤30 \boxed{18 \leq f(8) - f(2) \leq 30}

STEP 7

For problem 25, consider the derivative constraint f′(x)≤2 f'(x) \leq 2 .
The maximum possible change in f(x) f(x) from x=0 x = 0 to x=2 x = 2 is:
f(2)−f(0)≤2×(2−0) f(2) - f(0) \leq 2 \times (2 - 0)
f(2)−(−1)≤4 f(2) - (-1) \leq 4

STEP 8

Simplify the inequality:
f(2)+1≤4 f(2) + 1 \leq 4
f(2)≤3 f(2) \leq 3
Since f(2)=4 f(2) = 4 does not satisfy f(2)≤3 f(2) \leq 3 , such a function f f cannot exist.
The answers are:
23. 16 \boxed{16}
24. 18≤f(8)−f(2)≤30 \boxed{18 \leq f(8) - f(2) \leq 30}
25. No, such a function does not exist.

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