Question

Find the indicated values, where g(t)=t^2-t and f(x)=1+x f(2g(1)

Answer 1

$g(t)=t^2-t\\\\f(x)=1+x\\\\f(2g(1))=?\\\\g(1)=\\_{\text{put t=1 to the equation of g(t)}}\\\\=1^2-1=1-1=0\\---------------\\2g(1)=2(0)=0\\\\f(2g(1))=\\_{\text{put x-0 to the equation of f(x)}}\\\\=1+0=1\\\\Answer:\ f(2g(1))=1$

Answer 2

Answer:

The indicated value f(2g(1)) is: 1

Step-by-step explanation:

We are given two functions f(x) in terms of variable 'x' and g(t) in terms of variable 't' as:

                   $g(t)=t^2-t$

and          $f(x)=1+x$

Now we are asked to find the value of the composite function:

$f(2g(1))$

We know that:

$g(1)=1^2-1\\\\g(1)=0$

Hence,

$f(2g(1))=f(2\times 0)=f(0)\\\\\\f(0)=1+0\\\\\\f(0)=1$

                   Hence, the indicated value is:

                                     1