How to do this (1 Viewer)

[Sylfiphy]

I think u use second derivative but idk

 

[Unovan]

f'(x) = 2x + 1
f''(x) = 2

f''(a) = 2
f (a) = a^2 + a

f(a) = f''(a) when a^2 + a = 2

Now you can solve algebraically

 

[Sylfiphy]

f'(x) = 2x + 1
f''(x) = 2

f''(a) = 2
f (a) = a^2 + a

f(a) = f''(a) when a^2 + a = 2

Now you can solve algebraically

stop why was that easy thanks bro but also I have another question

 

[aqwerty13402]

stop why was that easy thanks bro but also I have another question

i dont even know what to do. im so cooked for hsc maths

 

[Unovan]

stop why was that easy thanks bro but also I have another question

If the rate of change (dV/dt) is positive (increasing, >0), the volume will be initially increasing. If the rate of change of the rate of change (d2V/dt2) is negative (decreasing, <0), the volume will begin decreasing in its rate of change. So you should first consider the initial rate of change, but then how that will change over time in your graph.

 

[Sylfiphy]

i dont even know what to do. im so cooked for hsc maths

legit me idk how this guy is even able to reply to me

 

[Sylfiphy]

If the rate of change (dV/dt) is positive (increasing, >0), the volume will be initially increasing. If the rate of change of the rate of change (d2V/dt2) is negative (decreasing, <0), the volume will begin decreasing in its rate of change. So you should first consider the initial rate of change, but then how that will change over time in your graph.

I GOT IT THANK U

 

[Unovan]

glad to hlep