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

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

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

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@tigel__ post the screenshot of the full problem please
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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?
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?
No I'm confused. on the first one
@tigel__ the one where you put -9700 ?
yes
(fg)(x) = (f*g)(x) (fg)(x) = f(x)*g(x) (fg)(4) = f(4)*g(4) (fg)(4) = -97*25 (fg)(4) = -2425
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)
Oh im sorry! I meant the other one the one above it,
oh ok
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
f(4) = -97 f ' (4) = -80 g ' (-97) = -390

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