Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A function g(t)g(t) increases by a factor of 77 over every unit interval in tt and g(0)=1g(0) = 1.\newlineWhich could be a function rule for g(t)g(t).\newlineChoices:\newline(A) g(t)=7tg(t) = -7t\newline(B) g(t)=7tg(t) = 7^t\newline(C) g(t)=17tg(t) = 1 - 7t\newline(D) g(t)=1.07tg(t) = 1.07^t

Full solution

Q. A function g(t)g(t) increases by a factor of 77 over every unit interval in tt and g(0)=1g(0) = 1.\newlineWhich could be a function rule for g(t)g(t).\newlineChoices:\newline(A) g(t)=7tg(t) = -7t\newline(B) g(t)=7tg(t) = 7^t\newline(C) g(t)=17tg(t) = 1 - 7t\newline(D) g(t)=1.07tg(t) = 1.07^t
  1. Check g(0)g(0): Check g(0)g(0) for each choice to see if it equals 11.\newline(A) g(0)=7(0)=0g(0) = -7(0) = 0, not 11.\newline(B) g(0)=70=1g(0) = 7^0 = 1, this is correct.\newline(C) g(0)=17(0)=1g(0) = 1 - 7(0) = 1, this is correct.\newline(D) g(0)=1.070=1g(0) = 1.07^0 = 1, this is correct.
  2. Check function increase: Now, check if the function increases by a factor of 77 over every unit interval in tt.\newline(A) is already incorrect.\newline(B) g(1)=71=7g(1) = 7^1 = 7, which is 77 times g(0)g(0).\newline(C) g(1)=17(1)=6g(1) = 1 - 7(1) = -6, which is not 77 times g(0)g(0).\newline(D) g(1)=1.071=1.07g(1) = 1.07^1 = 1.07, which is not 77 times g(0)g(0).
  3. Eliminate incorrect options: Eliminate the incorrect options based on the previous step.\newline(A) Incorrect.\newline(C) Incorrect.\newline(D) Incorrect.\newlineOnly (B)(B) remains as the possible correct answer.

More problems from Intermediate Value Theorem