Question

Find the Derivative of f(x)=2x^2 -x-3. a=1

Answer 1

Please show the step by step long method, not the power rule or whatever the shortcut is. They haven't taught us that yet, and we have to do it the tedious way for now. /

Answer 2

\[f'(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x}\]

Answer 3

That's the first principle, substituting in
\[f'(x)=\frac{2(x+\Delta x)^2-(x+\Delta x)-3-(2x^2-x-3)}{\Delta x}\]
Simplify

Answer 4

Yeah i did that. Except for Changeofx, i used h. Does it matter?

Answer 5

I get 2x+1 or something else, but the correct answer is 4x-1.

Answer 6

It doesn't

Answer 7

Hence why i'm asking the question, to show all steps please.

Answer 8

The correct answer is 4x-1. and the equation for the tangent line is y=3x-5. I don't see how.

Answer 9

The final answers are shown for the first 2, and the one with the green star for the graph is correct. I'm asking how to obtain these.

Answer 10

\[\frac{2(x+\Delta x)^2-x-\Delta x -3-2x^2 +x +3}{\Delta x}\]

Answer 11

Yup already got that. thanks

Answer 12

ok

Answer 13

Except i expanded the first part, is that wrong?

Answer 14

I have 2x^2 +2xh +h^2 then everything else you had.

Answer 15

You don't have to take the time make your responses perfect, although i appreciate it. I'm more just looking for the explanation.

Answer 16

\[\lim_{\Delta x \rightarrow 0}\frac{2(x^2+2x \Delta x + \Delta x^2)-2x^2-\Delta x}{\Delta x} \\ \\ \lim_{\Delta x \rightarrow 0} \frac{\cancel{2x^2}+4x \Delta x +2\Delta x ^2\cancel{-2x^2}-\Delta x}{\Delta x} \\ \\ \lim_{\Delta x \rightarrow 0} \frac{4x \Delta x+2\Delta x^2 -\Delta x}{\Delta x} \\ \\ \lim_{\Delta x \rightarrow 0}4x+2\Delta x +1\]
Then apply limit
you get
4x+1

Answer 17

Oh i see, but wait should't it be 4x+2h-1. Not plus 1.

Answer 18

The answer is 4x-1 anyways.

Answer 19

Which then simplifies to 4x-1, not 4x+1.

Answer 20

oh made a mistake there sorry, yeah its 4x-1

Answer 21

no problem. Also why is the limit 0? I thought a=1?

Answer 22

If you look at my screenshot, i don't see anywhere saying the limit is 0.

Answer 23

That is when f'(x)=a
and y will be f(a)

Answer 24

Whatever it works so i get it. Can you explain to me part B? what y= for the equation of the line tangent to the graph.

Answer 25

You gotta find the gradient first, we know that
\[f(x)=2x^2-x-3 \\ \\ f'(x)=4x-1\]
and the turning point is when f(1)
\[f'(1)=4-1=3\]

Answer 26

Okay, and then i use the forumla y-1=3(something goes here not sure what) Right?

Answer 27

Then, from the given turning point, (a,f(a)),
the y is f(a), so
back to the original equation,
\[y=f(a)=2a^2-a-3\]
when a=1,
\[f(1)=-2\]
So this is your y

Answer 28

Then use the point slope formula,
\[y+2=3(x-1)\]

Answer 29

Where coordinates are (1,-2) and m=3

Answer 30

Oh okay i think i got it.

Answer 31

good luck :)

Answer 32

I tried it on another problem, and i can't seem to get it.

Answer 33

f(x)=2x^2 +3x-2 when a=-2

Answer 34

The derivative is 4x+3, i got that right. But i can't figure out the y=. If i plug in -2 for a back into the original equation, it comes out to 0.

Answer 35

Which leaves me with y=3(x-2) which is 3x-6. But it is incorrect? :

Answer 36

you have the question paper?

Answer 37

Its the same as the last one.

Answer 38

Except f(x) = 2x^2 +3x-2. when a=-2

Answer 39

The derivative is 4x+3 i got it right. I just don't know how to solve for y the tangent line thing.

Answer 40

Turns out its y=-5x-10. flutter got it wrong again, gotta do the whole problem over again. What did i do wrong.

Answer 41

Is the actual answer y=-5x-10?

Answer 42

Stupid mistake, i think i get it now. I re read what you wrote, was a little confusing but figured it out.

Answer 43

okay

Answer 44

Thanks

Answer 45

yw