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
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OpenStudy (anonymous):
?
OpenStudy (anonymous):
\[f(x)=bx\] is a line with slope \(b\)
OpenStudy (anonymous):
unless perhaps you mean
\[f(x)=b^x\]
OpenStudy (dumbsearch2):
Sorry, my mistake. I meant that.
OpenStudy (anonymous):
thought that might be it
what are the conditions on the base so that an exponential function increases (grows) rather than decreases (decays) ?
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OpenStudy (astrophysics):
Sorry sat
OpenStudy (anonymous):
no problem, but that was not the answer in any case
OpenStudy (anonymous):
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
OpenStudy (astrophysics):
Yeah I realized that, since the functions isn't defined at b = 1, and b > 0.
OpenStudy (dumbsearch2):
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.?
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OpenStudy (astrophysics):
b<0* >_>, wow lol.
OpenStudy (anonymous):
yes
OpenStudy (dumbsearch2):
So would it be C. b > 1?
OpenStudy (anonymous):
the answer is always C
OpenStudy (astrophysics):
Yes, that's right! ^.^
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