g(x)=1x+x,g:R+→R+g(x)=\frac{1}{\sqrt{x + \sqrt{x}}}, \quad g: \mathbb{R}^+ \to \mathbb{R}^+g(x)=x+x1,g:R+→R+
Let f(x)=x,h(x)=x2+x, and k(x)=1x.f(x)=\sqrt{x}, h(x)=\sqrt{x^2+x}, \text{ and } k(x) = \frac{1}{x}.f(x)=x,h(x)=x2+x, and k(x)=x1.
Determine h∘f.h\circ f.h∘f.
h∘f=h(f(x))=(f(x))2+f(x)=(x)2+x=x+xh\circ f = h(f(x)) = \sqrt{(f(x))^2 + f(x)} = \sqrt{(\sqrt{x})^2 + \sqrt{x}} \\ \qquad = \sqrt{x + \sqrt{x}}h∘f=h(f(x))=(f(x))2+f(x)=(x)2+x=x+x
Now, determine k∘(h∘f).k\circ(h\circ f).k∘(h∘f).
k∘(h∘f)=k(h(f(x)))=1h(f(x))=1x+xk\circ(h\circ f) = k(h(f(x))) = \frac{1}{h(f(x))} = \frac{1}{\sqrt{x + \sqrt{x}} }k∘(h∘f)=k(h(f(x)))=h(f(x))1=x+x1
Therefore, g(x)=k∘(h∘f), where f(x)=x,h(x)=x2+x, and k(x)=1x.g(x) = k\circ(h\circ f), \text{ where } f(x)=\sqrt{x}, h(x)=\sqrt{x^2+x}, \text{ and } k(x) = \frac{1}{x}.g(x)=k∘(h∘f), where f(x)=x,h(x)=x2+x, and k(x)=x1.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments