Question

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

f(t) = t − 4 t^2, a = 3.

Answer 1

\[\begin{align} f'(t)&=\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}\\&=\lim_{h\to 0}\frac{t+h-4(t+h)^2-\left[ t-4t^2\right]}{h}\\
&=\lim_{h\to 0} \frac{t+h-4t^2-8th-4h^2-t+4t^2}{h}\\
&=\lim_{h \to 0}\frac{h-8th-4h^2}{h}\\
&=\lim_{h \to 0}\frac{h(1-8t-4h)}{h}\\&= \lim_{h \to 0}(1-8t-4h)\end{align} \]

Hopefully finding the limit of that is easy. Then, you just substitute for \(f'(a)\), since t=a and \(a=3\), substitute t=3 into the limit you found. This value will be the slope \(m\) of the tangent line \(f(t)=mt+b\) .

To find \(b\), you know that a=3, which is really t=3, so you can find \(f(3)\). (not \(f'(3)\)), and plug those into the line equation and solve for \(b\):
\(f(3)=m(3)+b\), you know \(m\) from the previous step.

Answer 2

@kirbykirby i got -35 for m

Answer 3

so is y=-35x+3?

Answer 4

not quite, actually when you compute the limit up there, you only set h to 0 in the formula, and let t alone. (so the 4h term will disappear as h goes to 0)

Answer 5

so what would y =?

Answer 6

y=-35x?

Answer 7

Nvm, so 1-8t?

Answer 8

\[ \lim_{h \to 0}(1-8t-4h)=1-8t\]
But now t=3 in your question, so \(1-8(3)=-23\), this is your slope \(m\).

Answer 9

so how would i figure out the equation?

Answer 10

(y-y1)=m(x-x1)?

Answer 11

so you now have \(y=-23t+b\) (I'm using t instead of x since they use t in your question).

And you can find the value \(b\) by using a point given. Since \(t=3\), you can find \(y=f(3)\) by plugging in t=3 into your original function

Or as you mention above, you can also use that form of the equation \((y-y_1)=m(t-t_1)\) and find \(y_1=f(3)\) and \(t_1=3\)

Answer 12

so y=-23t+35?

Answer 13

that's almost what I get: y=-23t+36

Answer 14

thank u so much!

Answer 15

your welcome :)

Answer 16

I sent you a message btw!