If g(x)=x^{2}+x-1 and (g circ f)(x)=4 x^{2}-1 | vedclass

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(B) Given $g(x) = x^{2} + x - 1$ and $(g \circ f)(x) = g(f(x)) = 4x^{2} - 10x + 5$. Let $f(x) = y$. Then $g(y) = y^{2} + y - 1 = 4x^{2} - 10x + 5$. $y

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) Given $g(x) = x^{2} + x - 1$ and $(g \circ f)(x) = g(f(x)) = 4x^{2} - 10x + 5$. Let $f(x) = y$. Then $g(y) = y^{2} + y - 1 = 4x^{2} - 10x + 5$. $y^{2} + y - 1 = 4x^{2} - 10x + 5$ $y^{2} + y - 6 = 4x^{2} - 10x$. To solve for $y$,we complete the square for $y^{2} + y$: $(y + \frac{1}{2})^{2} - \frac{1}{4} - 6 = 4x^{2} - 10x$. $(y + \frac{1}{2})^{2} = 4x^{2} - 10x + \frac{25}{4} = (2x - \frac{5}{2})^{2}$. Taking the square root,$y + \frac{1}{2} = \pm(2x - \frac{5}{2})$. Case $1$: $f(x) = 2x - \frac{5}{2} - \frac{1}{2} = 2x - 3$. Case $2$: $f(x) = -2x + \frac{5}{2} - \frac{1}{2} = -2x + 2$. For $f(x) = 2x - 3$,$f(\frac{5}{4}) = 2(\frac{5}{4}) - 3 = \frac{5}{2} - 3 = -\frac{1}{2}$. For $f(x) = -2x + 2$,$f(\frac{5}{4}) = -2(\frac{5}{4}) + 2 = -\frac{5}{2} + 2 = -\frac{1}{2}$. Thus,$f(\frac{5}{4}) = -\frac{1}{2}$.

If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5,$ then $f\left(\frac{5}{4}\right)$ is equal to

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