Asked by Farah
Let f(x)= (1/b)^x for b>1, and left g(x) = (f'(x))/(f(x))
a) Predict the shape of the graph of g(x).
it will be a straight line, since ln(1/b) will be a constant
b) Test your prediction by exploring two specific cases.
Values of (1/2) and (1/10)
c) Summarize your findings, using words and diagrams.
repeat part a) expect in words and graphs
d) For what value of b will g(x)= -1
it would be undefined, since there is no value of b, and you prove it by taking natural log on both sides and showing that ln -1 is a non real answer
please check the above answers.
a) Predict the shape of the graph of g(x).
it will be a straight line, since ln(1/b) will be a constant
b) Test your prediction by exploring two specific cases.
Values of (1/2) and (1/10)
c) Summarize your findings, using words and diagrams.
repeat part a) expect in words and graphs
d) For what value of b will g(x)= -1
it would be undefined, since there is no value of b, and you prove it by taking natural log on both sides and showing that ln -1 is a non real answer
please check the above answers.
Answers
Answered by
MathMate
(a) Correct.
Also, note that since the function is a constant, it does not contain x as a variable, therefore it is a horizontal line y=-log(b).
(d)For <i>what value of b</i> will g(x)= -1
g(x)=-ln(b)
so
-ln(b)=-1
ln(b)=1
raise to the power of e:
e^(ln(b)) = e^1
b=e (=2.718...)
Also, note that since the function is a constant, it does not contain x as a variable, therefore it is a horizontal line y=-log(b).
(d)For <i>what value of b</i> will g(x)= -1
g(x)=-ln(b)
so
-ln(b)=-1
ln(b)=1
raise to the power of e:
e^(ln(b)) = e^1
b=e (=2.718...)
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