## keneisha 3 years ago What is f(x+h) for the function f(x)=x^3+4x?

1. keneisha

that one is cube

2. hartnn

to get f(x+h), just replace all 'x' in f(x) by 'x+h'

3. keneisha

i don't understand

4. hartnn

example if f(x) = x^2+10 x f(x+h) will be (x+h)^2 + 10(x+h) note that i have replaced ALL 'x' in f(x) by x+h

5. hartnn

so, can you try replacing 'x' in f(x) by x+h for your question, x^3+4x ?

6. keneisha

x^3+4x+h

7. ParthKohli

\[f(\color{#C00}x) = \color{#C00}x^3 + 4\color{#c00}x \\ f(\color{#C00}h) = \color{#c00}h^3 + 4 \color{#C00}h \\ \vdots \]

8. keneisha

?????? i'm not sure

9. hartnn

wherever you 'see' x, just replace it by 'x+h' one more try , maybe ?

10. hartnn

u want 1 more example ?

11. keneisha

(x+h)^3+4(x+h)

12. hartnn

\(\checkmark \)

13. keneisha

is that right now?????

14. ParthKohli

That's it! Expand

15. hartnn

now, can you expand (x+h)^3 ?

16. hartnn

use \((a+b)^3 = a^3+3a^2b+3ab^2+b^3\)

17. ParthKohli

Just replace \(a\) by \(x\) and \(b\) by \(h\).

18. keneisha

x^3+2hx^2+2hx+hx^2+2h^2x+2h^2

19. ParthKohli

Uh, that is expanded a little too much.

20. keneisha

x^3+3hx^2+2h^x+2h^2+2hx

21. hartnn

where did h^3 go ?

22. keneisha

wheere will i get that h^3

23. hartnn

you'll have only 4 terms from that formula, right ?

24. ParthKohli

Four completely simplified terms.

25. hartnn

\((a+b)^3 = a^3+3a^2b+3ab^2+b^3\) from the last term b^3 of above

26. keneisha

i'm confused now

27. ParthKohli

Why?

28. keneisha

x^3+3x^h+3xh^2+h^3

29. hartnn

yes!

30. keneisha

is that right now. . .

31. hartnn

x^3+3x^2h+3xh^2+h^3

32. keneisha

then?

33. hartnn

so, you have (x+h)^3+4(x+h) =x^3+3x^2h+3xh^2+h^3 +4x+4h i think no further simplification possible, thats your final answer...

34. hartnn

(x+h) can be factored, but not required...

35. keneisha

thanks

36. hartnn

welcome ^_^