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gizmo_the_mogwai [7]
3 years ago
6

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:

25

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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Answer:

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Step-by-step explanation:

Assuming you are supposed to simplify:

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<h3>What are equivalent expressions?</h3>

Equivalent expressions are simply known as expressions with the same solution but different arrangement.

Given the index expressions;

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Using the exponent rule, the two values are have different bases but the same exponent and thus, we multiply the bases and leave the exponents the same way.

This can be written as;

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expand the bracket

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This is a topic on coordinate geometry. If you would like to find out more you can give my Instagram page a follow as I'll be updating on this topic soon. In addition, I will be posting other useful materials created by myself at the same platform too!

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