## anonymous one year ago How do you calculate new functions from old ones using table quantities (x, f(x), g(x), f'(x), g'(x))?

1. anonymous

Do you have a specific question? It might be easier to help you with more details.

2. anonymous

yes there's a table that goes along with it.

3. anonymous

yes there's a table that goes along with it.

4. jim_thompson5910

@tigel__ post the screenshot of the full problem please

5. anonymous

6. anonymous

The third one's the easiest by a slight margin. $\frac{f(-2)}{g(-2)+5}=\frac{23}{13+5}$ Do you see how I got that?

7. anonymous

The second one could come next. Recall that $$(fg)(x)=f(x)g(x)$$, so $(fg)(4)=f(4)g(4)=-97\times25$ For the other two, you'll need to take derivatives. By the product rule (for the fourth problem): $(fg)'(-2)=f'(-2)g(-2)+f(-2)g'(-2)=-32\times13+23\times(-10)$ By the chain rule (for the first one): $h'(4)=\bigg(g(f)\bigg)'(4)=g'(f(4))f'(4)=g'(-97)f'(4)=\cdots$ Does that make sense?

8. anonymous

No I'm confused. on the first one

9. jim_thompson5910

@tigel__ the one where you put -9700 ?

10. anonymous

yes

11. jim_thompson5910

(fg)(x) = (f*g)(x) (fg)(x) = f(x)*g(x) (fg)(4) = f(4)*g(4) (fg)(4) = -97*25 (fg)(4) = -2425

12. jim_thompson5910

I got f(4) and g(4) from the table look in the f(x) row and column that starts with 4 to get f(4) similar for g(4)

13. anonymous

Oh im sorry! I meant the other one the one above it,

14. jim_thompson5910

oh ok

15. jim_thompson5910

h(x) = g(f(x)) h ' (x) = g ' (f(x)) * f ' (x) ... chain rule h ' (4) = g ' (f(4)) * f ' (4) ... replace every x with 4 h ' (4) = g ' (-97) * f ' (4) ... use the table h ' (4) = (-390) * (-80) ... use the table h ' (4) = 31,200

16. jim_thompson5910

f(4) = -97 f ' (4) = -80 g ' (-97) = -390

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