Please help me rationalize the numerator.
$${\frac{\left(\left({\frac{{\mathtt{3}}}{{\sqrt{{\mathtt{x}}}}}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{h}}}}}}\right)\right)}{{\mathtt{h}}}}$$
The back of the book answer is $${\frac{{\mathtt{3}}}{\left({\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{h}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{h}}}}\right)}}$$
I have tried but cannot seem to get the answer. I believe I am having a simplifying problem, but if anyone can show me steps, that would be highly appriciated.
+23246 I would do it in two steps:
First: multiply the numerator and denominator by the common denominator:
(√(x + h))·(√x)
Giving:
[ 3√(x + h) + 3√x ] / [ h·√(x + h)·√x ]
and factor out the 3: 3[√(x + h) + √x ] / [ h·√(x + h)·√x ]
Now, multiply the numerator and denominator of this fraction by the conjugate of the numerator:
√(x + h) - √x
Numerator: the product of 3[√(x + h) + √x ] times [ √(x + h) - √x ]
= 3[ (x + h) - x ] = 3h
Denominator: the product of [ h·√(x + h)·√x ] times [ √(x + h) - √x ]
= h(x+h)√x - h√(x + h)·x = h[ (x+h)√x - √(x + h)·x ]
= h[ √x(x + h) - x√(x + h) ]
Cancelling the h from both the numerator and the denominator gives you their answer.
+23246 I would do it in two steps:
First: multiply the numerator and denominator by the common denominator:
(√(x + h))·(√x)
Giving:
[ 3√(x + h) + 3√x ] / [ h·√(x + h)·√x ]
and factor out the 3: 3[√(x + h) + √x ] / [ h·√(x + h)·√x ]
Now, multiply the numerator and denominator of this fraction by the conjugate of the numerator:
√(x + h) - √x
Numerator: the product of 3[√(x + h) + √x ] times [ √(x + h) - √x ]
= 3[ (x + h) - x ] = 3h
Denominator: the product of [ h·√(x + h)·√x ] times [ √(x + h) - √x ]
= h(x+h)√x - h√(x + h)·x = h[ (x+h)√x - √(x + h)·x ]
= h[ √x(x + h) - x√(x + h) ]
Cancelling the h from both the numerator and the denominator gives you their answer.
