operations with functions

operations with functions

Postby Guest » Sun Aug 25, 2019 8:12 am

Hi its a math problem. i cant remember the rules for adding , subtracting, multiplying and dividing fractions with roots . for example ,f(x) the square root of x+1 added to f (g) 2/the square root of x+1 . the answer was x+3 over the square root of x+1 but i dont know how they obtained this. the rules etc. thank you

p.s. i would like to know what topic to study concerning the rules of adding , multiplying , dividing root fractions. thank you
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Re: operations with functions

Postby Guest » Fri Aug 30, 2019 6:24 pm

The rules for working with functions are exactly the same as the rules for working with numbers because, for any function, f, f(x) is a number! Here, [tex]f(x)= \sqrt{x+ 1}[/tex] so, for example, [tex]f(3)= \sqrt{3+1}= \sqrt{4}= 2[/tex]. Similarly, with [tex]g(x)= \frac{2}{\sqrt{x+ 1}}[/tex] so that [tex]g(3)= \frac{2}{\sqrt{3+ 1}}= \frac{2}{2}= 1[/tex] so that [tex](f+ g)(3)= 2+ 1= 3[/tex].

More generally, [tex](f+ g)(x)= \sqrt{x+ 1}+ \frac{2}{\sqrt{x+ 1}}[/tex]. To add fractions, you need to have the same denominator. Here, there is only the denominator [tex]\sqrt{x+ 1}[/tex] so we need to multiply the first term by [tex]\frac{\sqrt{x+ 1}}{\sqrt{x+ 1}}[/tex]: [tex]\sqrt{x+ 1}\frac{\sqrt{x+1}}{\sqrt{x+1}}= \frac{x+ 1}{\sqrt{x+1}}[/tex]. Adding that to [tex]\frac{2}{\sqrt{x+ 1}}[/tex], [tex]\frac{x+1}{\sqrt{x+1}}+ \frac{2}{\sqrt{x+1}}= \frac{x+ 3}{\sqrt{x+ 1}}[/tex]. That is exactly what you would do if you were adding, say, [tex]1+ \frac{2}{3}= \frac{5}{3}[/tex].
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