Minimum value of function $f(x)=x+\log_2(2^{x+2}-5+2^{-x+2})$ out of 5 options
A : $\log_2(1/2)$
B : $\log_2(41/16)$
C : $39/16$
D : $\log_2(4.5)$
E : $\log_2(39/16)$
I just... don't know how to approach this.
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You could set $f'(x) = 0$ and solve for $x$ and see what you get, or you could fill in all the values and calculate which one gives the lowest value.. I would try the first method. – Nescrio Apr 16 '15 at 18:46
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What do you mean by setting $f′(x)=0$? – John Doe Apr 16 '15 at 18:49
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$f'(x)$ is the derivative. – Nescrio Apr 16 '15 at 18:51
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Welp, I need to solve the problem with a less complex mathematical apparatus, I don't know calculus (the problem isn't predicted for students that know calc) – John Doe Apr 16 '15 at 19:02
1 Answers
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$$x=\log_2(2^x)$$
$$f(x)=\log_2[2^x(2^{x+2}-5+2^{-x+2})]=\log_2[2^{2x+2}-5\cdot2^x+4]$$
Now $4\cdot2^{2x}-5\cdot2^x+4=(2^{x+1})^2-2\cdot2^{x+1}\cdot\dfrac54+\left(\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $=\left(2^{x+1}-\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $\ge4-\left(\dfrac54\right)^2$
lab bhattacharjee
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So the correct answer is $log_2(39/16)$?
Can I ask you how you got the idea to do what you did?
– John Doe Apr 16 '15 at 18:57 -
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But you ended up making one of $2^{x+1}$? Thanks! I tried something like this but didn't get the idea. – John Doe Apr 16 '15 at 19:17
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