## MTALHAHASSAN2 one year ago find the point on the curve f (x)=(x^2-1)^2 for which the tangent line is horizontal

1. IrishBoy123

$$f = (x^2-1)^2$$ you learning calculus?

2. MTALHAHASSAN2

yes

3. IrishBoy123

so you know to differentiate f(x) and set $$f'(x) = 0$$ ??

4. MTALHAHASSAN2

yeah i do

5. MTALHAHASSAN2

wait let be show you then

6. IrishBoy123

cool!

7. MTALHAHASSAN2

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8. MTALHAHASSAN2

right?

9. IrishBoy123

nah $$f = (x^2-1)^2$$ $$f' = 2(x^2-1)(2x)=\dots$$ if that is confusing, let $$u = x^2 -1$$ and use $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

10. MTALHAHASSAN2

then is it be 4x(x^2-1)

11. IrishBoy123

yes let's do it long way too $$f = (x^2-1)^2 = x^4 - 2 x^2 + 1$$ $$f' = 4x^3 - 4x = 4x(x^2-1)$$ $$\checkmark$$

12. IrishBoy123

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13. MTALHAHASSAN2

oh ok got you

14. MTALHAHASSAN2

thnx a lot so the 4x(x^2-1) be the tangent line is

15. IrishBoy123

$$f' = 0 \implies 4x(x^2-1) = 0$$ solve for x, you will get 3 turning points

16. IrishBoy123

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