Question

find the critical points of y=5x^3-6x​

Answer 1

Given :   y = 5x³ - 6x​  

To Find :   critical points

Solution:

y = 5x³ - 6x​  

dy/dx = 3(5x²) - 6

=> dy/dx = 15x²  - 6

dy/dx  = 0

=>  15x²  - 6 = 0

=>  x² = 6/15

=> x² = 2/5

=> x = ±√(2/5)

critical points are ±√(2/5)

d²y/dx² = 30x

x = √(2/5) => d²y/dx² > 0 Hence local minima

x = -√(2/5) => d²y/dx² <  0 Hence local maxima

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Answer 2

The critical points of $y=5x^3-6x$ are $\boxed{x=\sqrt{\frac{2}{5}}}$ and $\boxed{x=-\sqrt{\frac{2}{5}}}$

Step-by-step explanation:

Given:

$y=5x^3-6x$

To find out:

The critical points of the given function

Solution:

$y=5x^3-6x$

Taking the first derivative

$\frac{dy}{dx}=15x^2-6$

$\implies \frac{dy}{dx}=3(5x^2-2)$

Now a function f(x) has critical points where f'(x) = 0 or f'(x) does not exists

Therefore, for critical points

$\frac{dy}{dx}=0$

$\implies 3(5x^2-2)=0$

$\implies 5x^2-2=0$

$\implies x=\pm\sqrt{\frac{2}{5}}$

Therefore, the critical points of $y=5x^3-6x$ are $x=\sqrt{\frac{2}{5}}$ and $x=-\sqrt{\frac{2}{5}}$

Hope this answer is helpful.

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