@@turksvids but sir as f prime is in quadratic equation... So doesnt it mean that slope is always increasing??? .... Then how come f prime be negative? ... Sir plz clr my doubts... I shall be thankful to you 🙏
You have to be really careful with the words here. Let's say f(x) = x^2, so f'(x) = 2x, and f''(x)=2. In this case f'(x) changes negative to positive at x=0, so f(x) changes decreasing to increasing at x=0. It is also true that f''(x)=2>0 for all x, meaning that f'(x), the derivative of f(x) and the slope of f(x), is always increasing. So it is possible for the slope to be increasing while the slope changes from negative to positive. This video might be helpful: ruclips.net/video/CiCoMzZOBsk/видео.html I hope this helps!
The sign chart I used starting at around 2:29 shows a change in the first derivative from positive to negative at x = -2, so x = -2 is a relative maximum of the function.
Is there a way to do something like this for quartics? I got this question from my teacher: “Find the range of a for x⁴ - 4x² + 4 - a = 0 which has 4 real roots.” I tried using your method shown in this video which is extremely helpful for other questions but for that particular question, when I equate the derivative to zero, I got 0, √2i and - √2i. I'm stuck because I don't know what to do when there are complex numbers involved. I'd greatly appreciate it if you or anyone can help me with this!
Hi! I think maybe you just factored your derivative wrong? I got that f'(x) = 4x^3-8x = 4x(x^2-2) so f'(x) = 0 gives x = 0, x = sqrt(2) and x =-sqrt(2). A sign chart shows minimums at x = +/- sqrt(2) and a max at x = 0. If we force the max to be above zero and the min to be below zero, we'll have our range of answers. Hope this helps!
@@turksvids THANK YOU SO MUCHHH! Seriously, thank you! Silly me that's such a careless mistake 😅 I got this homework during the lockdown so I had to learn it by myself but I didn't know what to search for until, by pure luck, I found your video today (after 2 weeks of looking). I owe you so much 😭 Have a nice day!
thank you so so much i've been looking everywhere to find how to do this and you explained it so easily for me god bless you
Superb thank you so much respected sir..
You are brilliant sir.
Thank god you uploaded it
You are the best sir.
is there a way to solve this without using calculus
Wow🙏😇...... greatly appreciated 🙂
How can f prime be negative?
Everywhere a function, f(x), is decreasing the function's derivative, f'(x), will be negative and the reverse is true as well: if f'(x)
@@turksvids but sir as f prime is in quadratic equation... So doesnt it mean that slope is always increasing??? .... Then how come f prime be negative? ... Sir plz clr my doubts... I shall be thankful to you 🙏
You have to be really careful with the words here.
Let's say f(x) = x^2, so f'(x) = 2x, and f''(x)=2.
In this case f'(x) changes negative to positive at x=0, so f(x) changes decreasing to increasing at x=0.
It is also true that f''(x)=2>0 for all x, meaning that f'(x), the derivative of f(x) and the slope of f(x), is always increasing.
So it is possible for the slope to be increasing while the slope changes from negative to positive.
This video might be helpful: ruclips.net/video/CiCoMzZOBsk/видео.html
I hope this helps!
Thank you very nice concept
how we can say maxima at x= -2
The sign chart I used starting at around 2:29 shows a change in the first derivative from positive to negative at x = -2, so x = -2 is a relative maximum of the function.
@@turksvids and at x=1 graph negative to positive,what it means?
@@turksvids 2by3 also positive
thank you Sir
This is what i needed
Thanks
Is there a way to do something like this for quartics? I got this question from my teacher:
“Find the range of a for x⁴ - 4x² + 4 - a = 0 which has 4 real roots.”
I tried using your method shown in this video which is extremely helpful for other questions but for that particular question, when I equate the derivative to zero, I got 0, √2i and - √2i. I'm stuck because I don't know what to do when there are complex numbers involved. I'd greatly appreciate it if you or anyone can help me with this!
Hi! I think maybe you just factored your derivative wrong? I got that f'(x) = 4x^3-8x = 4x(x^2-2)
so f'(x) = 0 gives x = 0, x = sqrt(2) and x =-sqrt(2).
A sign chart shows minimums at x = +/- sqrt(2) and a max at x = 0.
If we force the max to be above zero and the min to be below zero, we'll have our range of answers. Hope this helps!
@@turksvids THANK YOU SO MUCHHH! Seriously, thank you! Silly me that's such a careless mistake 😅
I got this homework during the lockdown so I had to learn it by myself but I didn't know what to search for until, by pure luck, I found your video today (after 2 weeks of looking). I owe you so much 😭
Have a nice day!
@@turksvids and I'm sorry, I forgot to hit the subscribe button 🙏
Thank you so much