How do you find the critical points for #f(x)=8x^3+2x^25x+3#?
Hello,
Determine the derivative.
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To find the critical points of ( f(x) = 8x^3 + 2x^2  5x + 3 ), follow these steps:
 Compute the derivative of ( f(x) ) with respect to ( x ).
 Set the derivative equal to zero and solve for ( x ).
 The solutions obtained in step 2 are the critical points of the function.
Let's proceed with these steps:

Compute the derivative of ( f(x) ): [ f'(x) = 24x^2 + 4x  5 ]

Set the derivative equal to zero and solve for ( x ): [ 24x^2 + 4x  5 = 0 ]
This is a quadratic equation. You can use the quadratic formula to solve for ( x ): [ x = \frac{{b \pm \sqrt{{b^2  4ac}}}}{{2a}} ]
Where ( a = 24 ), ( b = 4 ), and ( c = 5 ). Substituting these values into the formula gives: [ x = \frac{{4 \pm \sqrt{{4^2  4(24)(5)}}}}{{2(24)}} ] [ x = \frac{{4 \pm \sqrt{{16 + 480}}}}{{48}} ] [ x = \frac{{4 \pm \sqrt{{496}}}}{{48}} ]
 Simplify and find the solutions for ( x ): [ x = \frac{{4 \pm 4\sqrt{{31}}}}{{48}} ] [ x = \frac{{1 \pm \sqrt{{31}}}}{{12}} ]
So, the critical points of the function ( f(x) = 8x^3 + 2x^2  5x + 3 ) are: [ x = \frac{{1 + \sqrt{{31}}}}{{12}} ] [ x = \frac{{1  \sqrt{{31}}}}{{12}} ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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