For what values of b is f(x) = bx an increasing function? A. Any real number B. b > 0 C. b > 1 D. 0< b <1
?
\[f(x)=bx\] is a line with slope \(b\)
unless perhaps you mean \[f(x)=b^x\]
Sorry, my mistake. I meant that.
thought that might be it what are the conditions on the base so that an exponential function increases (grows) rather than decreases (decays) ?
Sorry sat
no problem, but that was not the answer in any case
for an exponential function \[f(x)=b^x\] to be well defined it is always the case that \(b>0\) and also if \(b=1\) it is a constant the question asks for which \(b\) is \(b^x\)increasing
Yeah I realized that, since the functions isn't defined at b = 1, and b > 0.
Is it that if the base of the x variable is greater than 1, the exponential function is continuously increasing, and if its less than 1, the exponential function is continuously decreasing.?
b<0* >_>, wow lol.
yes
So would it be C. b > 1?
the answer is always C
Yes, that's right! ^.^
Thank you everyone :)
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