Question

Limit question is attached as a pic

Answer 1

anotehr question please help with

Answer 2

which one?

Answer 3

please help with both

Answer 4

@myininay which one are you doing?

Answer 5

\[f(x)=6x^2-2x\]
\[f(c)=6c^2-2c \]
\[f(c+h)=6(c+h)^2-2(c+h)=6(c^2+2ch+h^2)-2c-2h \]
\[f(c+h)=6c^2+12ch+6h^2-2c-2h \]
\[f'(c)=\lim_{h \rightarrow 0}\frac{f(c+h)-f(c)}{h}=\lim_{h \rightarrow 0} \frac{(6c^2+12ch+6h^2-2c-2h)-(6c^2-2c)}{h}\]
\[f'(c)=\lim_{h \rightarrow 0} \frac{12ch+6h^2-2h}{h}\]
\[f'(c)=\lim_{h \rightarrow 0}(12c+6h-2)=12c+6(0)-2=12c-2\]

Answer 6

wow!

Answer 7

now if you want f'(3)
replace c with 3

Answer 8

so is it 48

Answer 9

@sabrina, impressive huh? what is more impressive is that in a week you will be able to do this problem in your head using a quick and easy shortcut

Answer 10

but definitions are awesome

Answer 11

cool wat is teh shortcut

Answer 12

what is the slope of y=constant

Answer 13

0

Answer 14

The answer is 34.

Answer 15

0
we say y'=0 if y=constant

if \[y=6x^2 \text{ we say} y'=6(2)x^{2-1}=12x^{1}=12x\]
we can use the definition to prove the general case
\[y=x^n => y'=nx^{n-1}\]

Answer 16

we can this the power rule

Answer 17

for second one you will find that the slope of the tangent line is -27, which means the slope of the "normal" line (perpendicular line) will be
\[\frac{1}{27}\] since there is only one choice of lines with that slope, that is your answer

Answer 18

*call?

Answer 19

yep 36-2=34 not 48

Answer 20

oh sorry...
and @ satelite that is answer chice 3 right

Answer 21

yeah i guess. i didn't actually finish the problem. i just saw that there was only one line with slope
\[\frac{1}{27}\] and chose that one

Answer 22

wow that took no computation at all!! i wish i cud think that fast to save me time in tests