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Answer:
3/(√(3(x+h)) +√(3x))
Step-by-step explanation:
The conjugate of a binomial is the same sum, but with the sign changed. That is (a +b) and (a -b) are a conjugate pair: each is the conjugate of the other.
The purpose of multiplying expressions involving roots or complex numbers by their conjugate is to take advantage of the relation ...
(a +b)(a -b) = a² -b²
Here. you want to eliminate the h from under the radical, so squaring the radical containing h is useful for the purpose. Hence the conjugate gets involved.
"Multiply by the conjugate" means multiply both the numerator and the denominator by the conjugate. (Effectively, multiply by 1.)
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[tex]\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{3(x+h)}-\sqrt{3x}}{h}\\\\=\frac{(\sqrt{3(x+h)}-\sqrt{3x})(\sqrt{3(x+h)}+\sqrt{3x})}{h(\sqrt{3(x+h)}+\sqrt{3x})}=\frac{3(x+h)-3x}{h(\sqrt{3(x+h)}+\sqrt{3x})}\\\\=\frac{3h}{h(\sqrt{3(x+h)}+\sqrt{3x})}=\boxed{\frac{3}{\sqrt{3(x+h)}+\sqrt{3x}}}[/tex]
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Additional comment
In the end, you will want the limit as h → 0, which will be 3/(2√(3x)). You will notice that h=0 no longer makes the denominator zero, so the limit is found by simple evaluation.
When the "binomial" is a complex number of the form a+bi, its conjugate is a-bi, regardless of the sign of b. That is, the conjugate of a complex number is found by negating its imaginary part. The sign of the real part is left alone.
