find the minimum value of 2^x - 4x +1
@tanjung Do you know calculus?
yea, but still stuck
First find the derivative of this. Let this be y \[y=2^x-4x+1\] Find \(\frac{dy}{dx}\)
I want conform 2^x or x^2 ?
2^x not x^2
Yeah
from 2^x - 4x +1 you need to differentiate this there is a rule that the derivative of a^x = a^x * lna so let us apply this here to obtain f'(x) = 2^x * ln2 -4 and because we are finding critical calues we will let our slope equal zero therefor now we have 2^x * ln2 -4=0 and now we simply just solve fo x which is 2^x = 4/ln2 and to solve for x we would have to take ln on both sides to get ln2^x = 4/ln2 --> x*ln2 = 4/ln2 --> x = 4/(ln2)^2 :)
Not quite: \[f(x)=2^x-4x+1\] Solving for a local minimum: \[f^{\prime}(x)=2^xln({2})-4=0\] \[2^xln({2})=4\] \[2^x=\frac{4}{ln({2})}\] \[xln(2)=ln(\frac{4}{ln({2})})\] \[x=\frac{ln(\frac{4}{ln({2})})}{ln(2)}\]
oo thanks i forgot to include ln on the right side
That's one way where LaTeX makes it a lot easier to spot mistakes.
hoho..im wrong in used differential of 2^x, ... so, we need subtitute the value x (critical) to function original isn't it?
hey tanjung, are you here?
yeahhh
yeah, you have to substitute back the value of x in your original expression to find the minimum value
ya, igot it.. ymin=(4 - 7ln(2) +4*ln(ln(2))/ln(2).. value y not simple yea.. .
Did you match it?
yeahh, correct m if i still wrong, please
You found that \[x=\frac{ln(\frac{4}{ln({2})})}{ln(2)}\] so it's \[x=\log_2 (\frac{4}{\ln 2})\] since \[\frac{\ln a}{\ln b}=\log_b a\] Now we have \[2^x-4x+1\] Let's substitute \[2^{\log_2 (\frac{4}{\ln 2})}-4{\log_2 (\frac{4}{\ln 2})}+1\] we know that \[a^{\log_a b}=b\] so \[\frac{4}{\ln 2} -4\log_2 (\frac{4}{\ln 2})+1\] Now you can simplify this, can't you?
oke, thank you ash... i understand know.
good:D
haahh, sorri... forget to say thank you for valpey to,,, very" sorri sir :)
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