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3 years ago

If f(1) = 10 and f '(x) ≥ 3 for 1 ≤ x ≤ 6, how small can f(6) possibly be?

Mathematics

2 answers:

Hoochie [10]3 years ago

7 0

Answer:

Step-by-step explanation:

Our "best" case is when $f'(x) = 3$ everywhere - if it changes it means $f(x)$ will grow faster. But if $f'(x)$ is constant, it means $f(x)$ is the straight line [tex] y-10 = 3 (x-1) [tex], which passes through the point (6, 25)

Natalka [10]3 years ago

6 0

Answer:

$f'( x ) = \frac{f(b) - f(a)}{b - a} \\ = \frac{f(6) -f(a)}{6 - 1} \\ 3\leqslant \frac{f(6) - 10}{5} \\ 15\leqslant f(6) - 10 \\ f(6) \geqslant 15 + 10 \\ f(6) \geqslant 25$

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13 percent of the drivers stopped for speeding have invalid licenses.

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(b) Exactly one will have an invalid license.

This is $P(X = 1)$

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(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

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$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56$

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