If f(1) = 13 and f '(x) ≥ 3 for 1 ≤ x ≤ 3, how small can f(3) possibly be?
1 answer:
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Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
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If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
Answer:
(a) 0.2707
(b) 0.8576
Step-by-step explanation:
