Let $f :$ $R \to R$ be given by $f (x) = x^{2} + 3 .$ Find ( a ) ${x : f (x) = 28}$ ( b ) the pre-images of $39$ and $2$ under $f .$

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Solution

We have function R $\to$ R defined by f(x) $= x^{2} + 3$
i.e its domain & range are real no.
1) f(x) = 28 = $x^{2} + 3$
$x^{2} = 25 \Rightarrow x = \pm 5$
Therefore , for range 28 Pre - image is 5 & -5
2) f(x) = 39.
$x^{2} + 3 = 39$
$x^{2} = 36 \Rightarrow x = \pm 6$
Now f(x) = 2
$x^{2} + 3 = 2$
$x^{2} = - 1$
$x = \pm i$ But the value of x is not are
value, both i & -i are complex no.
therefore, 2 is not an image under f.

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Let f: R→R be given by f(x)=x2+3. Find ( a ) {x:f(x)=28} ( b ) the pre-images of 39 and 2 under f.

Let $f :$ $R \to R$ be given by $f (x) = x^{2} + 3 .$ Find ( a ) ${x : f (x) = 28}$ ( b ) the pre-images of $39$ and $2$ under $f .$