The Student Room Group

Maths

Find the values of x for which f(x) is an increasing function, given that f(x)=5x^3+12x

That answer is that x belongs to real numbers.

I dont understand how you do this question. Ive found the derivative but i dont know what to do next
Original post by Anonymous202
Find the values of x for which f(x) is an increasing function, given that f(x)=5x^3+12x

That answer is that x belongs to real numbers.

I dont understand how you do this question. Ive found the derivative but i dont know what to do next


If the function is increasing, we have that f(x)0f'(x) \geq 0.

So you want to solve the inequality f(x)0f'(x) \geq 0.
Original post by RDKGames
If the function is increasing, we have that f(x)0f'(x) \geq 0.

So you want to solve the inequality f(x)0f'(x) \geq 0.

Thank ypu very much for the reply. I tried to do that amd i couldnt solve it. Id really appreciate it if you solved it for me.
Original post by Anonymous202
Thank ypu very much for the reply. I tried to do that amd i couldnt solve it. Id really appreciate it if you solved it for me.


Can't do that, but I don't see why you're struggling with this.


f(x)=5x3+12x    f(x)=15x2+12f(x) = 5x^3+12x \implies f'(x) = 15x^2 + 12.

So solving f(x)0f'(x) \geq 0 is the same as solving 15x2+12015x^2 + 12 \geq 0.

Divide through by 3 and you get 5x2+405x^2 + 4 \geq 0.


So, what are the possible values of xx ? I mean, you got the answer already, but do you see *why* it's the answer?
Hi, just passing by. Would you need to find the second derivative of the equation to answer this?

Typically, I like to draw a brief sketch of the curve itself to get an idea of what I'm expecting to get. :P
Original post by RDKGames
If the function is increasing, we have that f(x)0f'(x) \geq 0.

So you want to solve the inequality f(x)0f'(x) \geq 0.
Original post by KJEKJE Says Hi
Hi, just passing by. Would you need to find the second derivative of the equation to answer this?

Typically, I like to draw a brief sketch of the curve itself to get an idea of what I'm expecting to get. :P


Nope, you don't need the second derivative here.
Second Derivate is to find whether it is a maximum minimum or point of inflextion.
Original post by RDKGames
Nope, you don't need the second derivative here.
Original post by Anonymous202
i think i get it now. Is the answer because all values for x^2 are positive?


Yep, and from there you can quite simply show how 5x2+405x^2 + 4 \geq 0 must be true for all xRx \in \mathbb{R}.
Original post by RDKGames
Nope, you don't need the second derivative here.


Oh shoot, wait. Yeah I got it. Didn't notice that the equation passed through (0,0). ^^
Original post by KJEKJE Says Hi
Oh shoot, wait. Yeah I got it. Didn't notice that the equation passed through (0,0). ^^


Cool, but I don't see why (0,0) is significant to you here. It shouldn't be :smile:
Thank you so much!!!! Ive been stuck on this question for such a long time now. Thank you all the help.
Original post by RDKGames
Yep, and from there you can quite simply show how 5x2+405x^2 + 4 \geq 0 must be true for all xRx \in \mathbb{R}.
Isn’t is just asking for the places where the derivative is bigger than 0
Original post by RDKGames
Cool, but I don't see why (0,0) is significant to you here. It shouldn't be :smile:

Oh, I said that because when you found the derivative, it was a positive quadratic. I just thought that because you knew that the cubic equation passed through (0,0), all of the points bigger than 0 would be increasing in gradient.

Quick Reply

Latest