Answer:
option (b) is correct.
The value of [tex]g(f(3)-2f(1))[/tex] is 0.
Step-by-step explanation:
Given: [tex]g(t)=t^2-t[/tex] and [tex]f(x)=1+x[/tex]
We have to find the value of [tex]g(f(3)-2f(1))[/tex]
Consider the given function [tex]f(x)=1+x[/tex]
First evaluate value of function f(x) at x = 3 and 1 then substitute it in [tex]g(f(3)-2f(1))[/tex] .
f(3) = 1 + 3 = 4
f(1) = 1 + 1 = 2
Substitute, we get
[tex]g(f(3)-2f(1))=g(4-2(2))[/tex]
Solving further , we get,
[tex]g(f(3)-2f(1))=g(4-4)=g(0)[/tex]
Now evaluate [tex]g(t)=t^2-t[/tex] at t = 0 , we get,
[tex]g(0)=0^2-0=0-0=0[/tex]
Thus, value of [tex]g(f(3)-2f(1))[/tex] is 0.
Thus, option (b) is correct.
Answer:
b. 0
Step-by-step explanation:
Given: g(t) = t^2 - t and f(x) = 1 + x
Let's find f(3) and f(1)
f(3) = 1 + 3 = 4
f(1) = 1 + 1 = 2
Now we have to plug in these values, we get
g(f(3) - 2f(1)) = g(4 - 2(2))
= g(4 - 4)
= g(0)
g(0) = 0^2 - 0
g(0) = 0
The value of is b. 0
Hope you will understand the concept.
Thank you.