calculus

[lean terrace]

Here the differentiate will be 2x+a>0

[delicate pier]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[sacred gull]

If I understand the question correctly, It is asking for the min(a) for which
$(f'(x) \geq 0) for (x \in [0,1])$

[fossil fjord]

muerte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sacred gull]

So $2x+a \geq 0 for x \in[0,1]$
Since in the case of x = 0, it would amount to $a \geq 0$, a = 0 is the smallest possible a that satisfies the condition

[fossil fjord]

muerte

[lean terrace]

@sacred gull

[sacred gull]

for x 1,2

I thought its x 0,1

mb

If it is x [1,2] you have $2x+a \geq 0$ and if you insert x = 1 you get: $2 + a \geq 0$, which gives you $a \geq -2$

[fossil fjord]

muerte

[sacred gull]

I read the wrong intervall 😄

[lean terrace]

ohh it is okay

we just need to plug in 1

i see

thanks

[sacred gull]

ye

In questions like this, its really usefull to just input the lower bound

since you just need the smallest a possible

we just look at the edge case

[lean terrace]

if they have asked us to max value

then we would have use 2

instead of 1

[sacred gull]

Well if they ask you for the max value, you need to use whatever satisfies the equation

In this case, a could be infinitely high lol

[lean terrace]

yes right

thanks

[sacred gull]

no problem 🙂

[lean terrace]

you joined today?

[woven relic]

what does increasing mean here?

[lean terrace]

dy/dx>0

[woven relic]

ok

so we require, 2x + a >= 0 while x in [1,2]. Rewrite it as : a >= -2x as x in [1,2] we see that all a>= -2 will work. Thus it's opt (d)

[lean terrace]

yes bhut hi nice explanation hain bhai