If $f (x + 1) = 3 x - 9$ then find the value of $f (x^{2} - 1)$ .

Open in App

Solution

Given,
$f (x + 1) = 3 x - 9$ ......(1)
Now putting $x^{2} - 2$ in (1) in the place of $x$ then we get,
$f (x^{2} - 1) = 3 (x^{2} - 2) - 9$
or, $f (x^{2} - 1) = 3 x^{2} - 15$ .

Suggest Corrections

Similar questions

f (x + 1) = 3 x - 9

f (x^{2} - 1)

f (x) = x^{2} - 3 x + 4

x_{1} + x_{2}

x_{1}, x_{2}

f (x) = f (2 x + 1) .

f (x) = 3 - 2 x

x

f (x^{2}) = [f (x)]^{2}

f (x^{2}) f (y^{2}) - \frac{1}{2} [f (\frac{x^{2}}{y^{2}}) + f (x^{2} y^{2})]

f (x) = {log}_{0.1} (\frac{1 - | x |}{1 + | x |}),

x,

f (x) \geq f (x^{2})

Join BYJU'S Learning Program