Question

10. Let f: R --- R be given by f(x)= x² + 3. Find
(1) {x: f(x)=28}

(ii) The pre- images of 39 and 2 under 'f'.
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Answer 1

Answer:

Step-by-step explanation:

i)

x^2 + 3 =28

x^2 = 25

x = +- 5

ii)

pre- images of 39 and 2

f(x) = 39

x^2 + 3 =39

x^2 = 36

x = +-6

f(x) = 2

x^2 + 3 = 2

x^2 = -1

not possible

x -> Ф

Hope it helps :-)

Answer 2

Answer:

1) x=5 for f(x) = 28

2) x = +-6 for f(x) = 39 and x not possible for f(x) = 2

Step-by-step explanation:

1. It is given to us that f(x) = 28
   $f(x) = 28\\x^2+3=28\\x^2 = 25\\x=5$
2. 1. It is given to us that f(x) = 39
       $f(x) = 39\\x^2+3 =39\\x^2=36\\x=6$
   2. But for f(x) = 2
        $f(x)=2\\x^2+3=2\\x^2=-1$
       Not possible
   #SPJ3