Calculus I Simplifying and Substitution

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For the first one, I evaluated by replacing x for 2+h for f(x). Then I substituted into the givern expression and simplified to get (6h-5x)/(xh+h^2 -h)

For the second one, I replaced -1 for x in f(x) and got 2. Then I substituted into the given expression to get [√ (1-3x) - 2 ]/(x+1)

Any ideas on what to do?

Thanks in advance.

 

[MarkFL]

1.) Given:

$$f(x)=\frac{x+3}{x-1}$$

then:

$$f(2+h)=\frac{(2+h)+3}{(2+h)-1}=\frac{h+5}{h+1}$$

and

$$f(2)=\frac{2+3}{2-1}=\frac{5}{1}=5$$

and so:

$$\frac{f(2+h)-f(2)}{h}=\frac{\dfrac{h+5}{h+1}-5}{h}$$

Now, your task is to simplify...

 

[ardentmed]

1.) Given:

$$f(x)=\frac{x+3}{x-1}$$

then:

$$f(2+h)=\frac{(2+h)+3}{(2+h)-1}=\frac{h+5}{h+1}$$

and

$$f(2)=\frac{2+3}{2-1}=\frac{5}{1}=5$$

and so:

$$\frac{f(2+h)-f(2)}{h}=\frac{\dfrac{h+5}{h+1}-5}{h}$$

Now, your task is to simplify...

Alright, I simplified the expression and crossed out the extra h in the nominator and denominator and ultimately computed:

-4 / (h+1)

Is that right? How about the second one?

Thanks again.

 

[MarkFL]

Alright, I simplified the expression and crossed out the extra h in the nominator and denominator and ultimately computed:

-4 / (h+1)

Is that right? How about the second one?

Thanks again.

Yes, that's correct.

For the second one, you need to rationalize the numerator, which means multiplying by the conjugate...can you state the conjugate?