Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@tigel__ post the screenshot of the full problem please

Answer 5

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?

Answer 6

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?

Answer 7

No I'm confused. on the first one

Answer 8

@tigel__ the one where you put -9700 ?

Answer 9

yes

Answer 10

(fg)(x) = (f*g)(x)

(fg)(x) = f(x)*g(x)

(fg)(4) = f(4)*g(4)

(fg)(4) = -97*25

(fg)(4) = -2425

Answer 11

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)

Answer 12

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

Answer 13

oh ok

Answer 14

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

Answer 15

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