Given the exponential function A(x) = P(1 + r)x, what value for r will make the function a growth function? r = −2.1 r = −0.1 r = 0 r = 0.1
A(x)=P(1+r)\(^x\) like this ?
yeah
so your rate is 1+r. Why? because: P(1+r)\(^1=\)P(1+r) P(1+r)\(^2=\)P(1+r)(1+r) P(1+r)\(^3=\)P(1+r)(1+r)(1+r) P(1+r)\(^4=\)P(1+r)(1+r)(1+r)(1+r) and so on....
so you always multiply times another 1+r if 1+r=1, then you multiply just times 1 and value doesn't change, if 1+r is less than 1, then your value is going to decrease (because if you multiply times 0.6 9/10 or any number smaller than 1) you make the value smaller. if 1+r is greater than 1, [only] then will your value be growing
So, for what value of r (the value of r in which answer choice), gives you a "1+r" that is GREATER than 1?
ohh so 0.1
Yes, correct r=0.1 is the answer
thank you so much medal and fan
no need, if you want to I won't refuse xD, but you aren;t required:)
Good work!
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