Question

Use the limit definition to compute f′(a) and find an equation of the tangent line.

f(t) = t − 10 t^2, a = 3.

Y=

Answer 1

What limit definition would be to your liking?

Answer 2

Umm the easiest one! (:

Answer 3

doesnt it have to be f(x+x)-f(x)/x

Answer 4

I'm sure you mean "f(x+h)" and "/h". I'm also sure you mean \(\dfrac{f(x+h) - f(x)}{h}\), but this is not what you have written. Given the first two corrections, you have written \(f(x+h) - \dfrac{f(x)}{h}\), and this is absolutely not what you intend. Please remember your Order of Operations!

Okay, having said that, you'll need:

f(t) - which you have.
f(t+h) - Please show this in it's expanded form.

Answer 5

Okay, make that f(3) and f(3+h).

Answer 6

Yea I meant the top equation, what do you mean by expanded form?

Answer 7

f(3):(3-10(3)^2

Answer 8

f(3) = 3 - 10(3)^2 = 3 - 10(9) = 3 - 90 = -87

f(3+h) = (3+h) - 10(3+h)^2 = 3 + h - 10(9 - 6h + h^2) = 3 + h - 90 + 60h - 10h^2
f(3+h) = -87 + 61h - 10h^2

Okay, now assemble the numerator and simplify.

Answer 9

10(3)^2-61(3)-87=6

Answer 10

or do i do the derivative and then plug in?

Answer 11

I don't know what that is.

f(3+h) - f(3) = ( -87 + 61h - 10h^2) - (-87)

Simplify.

Answer 12

f(3+h)-f(3)= 7569+5407h-870h^2
or is it
61h-10h^2?

Answer 13

I am not encouraged by that. You should know the difference between multiplication and subtraction. It is the latter.

Okay, now we have the numerator, let's go get the denominator and simplify again.

Answer 14

Sorry my computer would not load, am still confused on what to do.

Answer 15

should be
\[f'(a)=\lim_{x \rightarrow a}\frac{ f(x-a)-f(a) }{ x-a }\]

Answer 16

would it be, (t-10t^2-3)-(3)/(t − 10 t^2-3)?

Answer 17

\[\lim_{t \rightarrow 3}\frac{ f(t-3)-f(3) }{ t-3 }\]

Answer 18

what's f(t-3) = ?
what's f(3) = ?

figure these out and plug them in. then you'll need to factor out (t-3)

Answer 19

What do I plug in for t?

Answer 20

in the first one?

Answer 21

\[f(t-3)=(t-3)-10(t-3)^2\]

Answer 22

I got an answer of 3

Answer 23

or do you want -10 t^3+90 t^2-270 t+270

Answer 24

sorry it's like this...
\[f'(a)=\lim_{x \rightarrow a}\frac{ f(x)-f(a)}{ x-a }\]

so we get...\[f'(3)=\lim_{x \rightarrow 3}\frac{ (x-10x^2)-(3-10(3)^2) }{x-3 }=\lim_{x \rightarrow 3}\frac{ (x-10x^2)+87 }{x-3 }\]
\[=\lim_{x \rightarrow 3}\frac{ (x-3)(-10x-29) }{x-3 }==\lim_{x \rightarrow 3}-10x-29=-59\]

Answer 25

ok I see what you did, however when i type in -59 as the answer in the website it comes up at incorrect?

Answer 26

hold on, perhaps I miscalculated...
nope that's it.

f(t) = t-10t^2, f'(t) = 1-20t... f'(3) = 1 - 20(3) = 1 - 60 = -59

Answer 27

that's just the slope of the tangent line... it's not the equation.
now that you have the slope, it has to go through the point (3, -87). use these facts to write the equation of the line with slope = -59, going through the point (3, -87)

Answer 28

oh thats it I have to use an equation y=-59x-87 right

Answer 29

no... when x = 3 y = -87... that's not the y intercept. this is calculus. you should have linear equations mastered!

Answer 30

lol hold on let me try this again

Answer 31

y+87=-59(x-3)

Answer 32

y=-146x+177

Answer 33

grrr now it says this Entered Answer Preview Result Messages
-146x+177 incorrect Variable 'x' is not defined in this context

Answer 34

the slope is wrong anyways... it's not -146, it's -59. as for how the system takes it, I can't help you there.

Answer 35

o i finally get why it was wrong instead of x its t, thanks for all the help!