Functions and Relations.(If f(x + 1) = x² + 4x - 5, find f(x - 2)

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[@yosualeonardo7443]

Nice explanation! I don't know if my way is correct but here is what I do: f(x+1) = x²+4x-5 f(x+1) = (x+5)(x-1) f(x+1) = (x+1+4)(x+1-2) f(x-2) = (x-2+4)(x-2-2) f(x-2) = (x+2)(x-4) f(x-2) = x²-2x-8

[@mathphobia2560]

It's absolutely correct Sir... Very nice approach 👍 Please do well to subscribe for more videos 🙏🏿

[@mathphobia2560]

Please subscribe to my channel for more videos 🙏🏻🙏🙏🏼🙏🏽🙏🏿

[@myelectronicsworld]

You could have set x=u-3 from the very beginning and you get the same result.

[@mathphobia2560]

That's very true Sir. I wanted to do it in this way because I thought this is a more easier way to understand. Nevertheless, your approach is also acceptable 👌

[@seasea5938]

@@mathphobia2560。我的看法是 一般學生應該比較容易理解Electronics的解法，利用參數直接使用已知函數代入才是更直接的做法。

[@mathphobia2560]

@@seasea5938 That's a very nice opinion Sir. I've made another video under this same topic and I explained this concept in that video. Thank you for your opinion tho. Very much appreciated.

[@Caturiya]

Very simple put x= u- 3. Then replace u by x. He must know this to, so he wanted to give us a possibility to be better than the teacher as antiarrogance support

[@amitavadasgupta6985]

My explanation shall b from my methodical correct logic not from some one else.

[@aitheria6804]

Good video with good explanation. Thanks man i love it ❤

[@mathphobia2560]

I'm glad you loved it bro. Please do well to subscribe for more videos 🙏🏼🙏🏼🙏🏼

[@samkay6430]

Very nice explanation. Thank you so much...

[@mathphobia2560]

Thank you 😊. My first ever comment I'm getting!

[@eswyatt]

So much of math is procedure (Fourier Series, solving canonical PDEs, . . .). What's happening here is really thinking!!!

[@mathphobia2560]

Yep... Many people really prefer working with formulas to the method of reasoning... But they all give the same results

[@amitavadasgupta6985]

f(x-2)=(x-3)^2+4(x-3)-5=x^2-2x-8 ans got by replacing x by x -3

[@mathphobia2560]

Exactly. 💯

[@amitavadasgupta6985]

f(x-2)=(x-3)^2+4(x-3)-5=x^2-2x-8

[@amitavadasgupta6985]

f(x-2)=x^2-2x-8 replacing x by x-3

[@mathphobia2560]

Please what are you trying to say?

[@MrLidless]

f(x) = ax^2 + bx + c f(x + 1) = ax^2 + (2a + b)x + (a + b + c) = x^2 + 4x - 5 a = 1, b = 2, c = -8 f(x) = (x + 4)(x - 2) f(x - 2) = (x + 2)(x - 4)

[@mathphobia2560]

Very cool method Sir... Indeed there are a lot of ways to kill a cat

[@deltalima6703]

f(x-2) = x² - 2x - 8

[@samuelweir5985]

I would have just noted that if you make the substitution x->x-3 on both sides of f(x+1)=x^2+4x-5, then f(x+1) becomes f(x-2).

[@mathphobia2560]

Very true Sir... I just wanted to use a different kind 9f substitution and that's why I solved it like that.... Nevertheless, what you're saying is very correct...

[@hamoudiayoub9341]

f(x-2) = f((x-3)+1) = (x-3)² + 4(x-3) - 5 = x² - 2x - 8

[@mathphobia2560]

Very nice one

[@NurHadi-qf9kl]

.x-1=x+1-3 .f=(x-3)^2+4(x-3)-5=x^2-2x-3

[@mathphobia2560]

Good. But your simplification isn't well done. (x-3)² + 4(x-3) - 5 = x²-6x+9+4x-12-5 = x² - 2x - 8

[@user-fq4gu3gm9z]

f(x + 1) = x² + 4x - 5, f(x + 1) = （x+1）² + 2(x +1)- 8,,,so f(x-2)=（x-2）² + 2(x -2)- 8=x² -4x+4+2x-4-8=x² -2x-8

[@mathphobia2560]

👍