Question

given f(x)= 4x + 1/4 and g(x) = root x^3, them find (gof^-1)(3/8) .​

Answer 1

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{128\sqrt{2} }$ or $\frac{\sqrt{2} }{256}$

Given:

f(x) = 4x + $\frac{1}{4}$

g(x) = √x³

To Find:

(gof⁻¹)($\frac{3}{8}$) =?

Solution:

f(x) = 4x + $\frac{1}{4}$

Let f(x) = y

So, y = 4x + $\frac{1}{4}$

y - $\frac{1}{4}$ = 4x

x = $\frac{y - \frac{1}{4} }{4}$

∴ f⁻¹(x) = $\frac{x - \frac{1}{4} }{4}$

g{f⁻¹(x)} = g( $\frac{x - \frac{1}{4} }{4}$)

g{f⁻¹(x)} = $\sqrt{(\frac{x - \frac{1}{4} }{4})^{3} }$

(gof⁻¹)($\frac{3}{8}$) = $\sqrt{(\frac{\frac{3}{8} - \frac{1}{4} }{4})^{3} }$

(gof⁻¹)($\frac{3}{8}$) = $\sqrt{(\frac{\frac{1}{8}}{4})^{3} }$

(gof⁻¹)($\frac{3}{8}$) = $\sqrt({\frac{1}{32}) ^{3} }$

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{32} ^{\frac{3}{2} }$

(gof⁻¹)($\frac{3}{8}$) = $(\frac{1}{32} *\frac{1}{32} *\frac{1}{32} )^{}\frac{1}{2}$

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{32} * (\frac{1}{32} )^\frac{1}{2}$

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{32} *\frac{1}{4} * \frac{1}{\sqrt{2} }$

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{128\sqrt{2} }$

(gof⁻¹)($\frac{3}{8}$) = $\frac{\sqrt{2} }{256}$

(gof⁻¹)($\frac{3}{8}$) = $\frac{1}{128\sqrt{2} }$ or $\frac{\sqrt{2} }{256}$

#SPJ1