Stuck on how to do: given f(x)= 2x^2-5x+1 simplify f(x+5)-f(x)/5

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ns10239

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1. given f(x)= 2x^2-5x+1 simplify f(x+5)-f(x)/5 The whole expression f(x+5)-f(x) is over 5 not just the f(x). The answer is 4x+5.

2. given f(x)= x^2-2x+5 and h(x)= x-2/3x-2 simplify f[h(x)]. the answer is 25x^2-78x+61/(3-2x)^2 the whole expression 25x^2-78x+61 is over (3-2x)^2 not just 61.

3.(5x^-2/3+x^-4/2)(x^1/6) simplify exponents, -2/3,-4/2, and 1/6 are fraction exponents. answer is 5√ x/x+1/x^11/16. 11/16 is a fraction exponent, 5√ x is over x.

thank you.
 
First, you seem to be aware that f(x+5)-f(x)/5 is not the same as (f(x+5)-f(x))/5. Just add the parentheses. No need to explain it.
Second, how does function notation work? If I give you h(x) = 3x-4, what is h(2)?
 
2, so far I got:

[2(x+5)^2-5(x+5)+1-2x^2-5x+1] /5,

[2(2x+10x+25)-5x-5+1-2x^2-5x+1]/5

[4x+20x+50-5x-5+1-2x^2-5x+1]/5

[-2x^2 -14x +57]/5 this is where Im stuck
 
ns10239 said:
3.(5x^-2/3+x^-4/2)(x^1/6) simplify exponents, -2/3,-4/2, and 1/6 are fraction exponents. answer is 5√ x/x+1/x^11/16. 11/16 is a fraction exponent, 5√ x is over x.
Click to expand...

Type grouping symbols around your fractional exponents. Use other grouping symbols for clarity is a recommendation. Space out your characters for more readability.
(I don't need the asterisk between the brackets below, but it was placed for emphasis.)


[5x^(-2/3) + x^(-4/2)]*[(x^(1/6)] . . . . . . . . . . . . . . . . . . . . . . . . The answer is 5√x/x + 1/x^(11/16).


This can get you started:


\(\displaystyle 5x^{-2/3}(x^{1/6}) \ + \ x^{-4/2}(x^{1/6}) \ = \)

\(\displaystyle 5x^{-2/3 \ + \ 1/6} \ + \ x^{-4/2 \ + \ 1/6} \ = \)


Please show your work from here to see if you can finish it.
 
Last edited:
lookagain said:
Type grouping symbols around your fractional exponents. Use other grouping symbols for clarity is a recommendation. Space out your characters for more readability.
(I don't need the asterisk between the brackets below, but it was placed for emphasis.)


[5x^(-2/3) + x^(-4/2)]*[(x^(1/6)] . . . . . . . . . . . . . . . . . . . . . . . . The answer is 5√x/x + 1/x^(11/16).


This can get you started:


\(\displaystyle 5x^{-2/3}(x^{1/6}) \ + \ x^{-4/2}(x^{1/6}) \ = \)

\(\displaystyle 5x^{-2/3 \ + \ 1/6} \ + \ x^{-4/2 \ + \ 1/6} \ = \)


Please show your work from here to see if you can finish it.
Click to expand...

5x^(-3/6) + x^(-11/6)

5x^(-1/2) + 1/x(11/6)

1/5x^(1/2) + 1/x(11/6)????
 
ns10239 said:
5x^(-3/6) + x^(-11/6)

5x^(-1/2) + 1/x(11/6) You're missing a "^" symbol.

1/5x^(1/2) + 1/x(11/6)????]The negative exponent is not operating on the 5, and you're still missing the "^" symbol.
Click to expand...

5x^(-3/6) + x^(-11/6)

5x^(-1/2) + x^(-11/6) . . . . I recommend you leaving it in this form. However, check with the instructions/instructor
to see if negative exponents are allowed in the final answer.
 
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