How do you find local maximum value of f using the first and second derivative tests: #f(x) = 4x^3 + 3x^2  6x + 1#?
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To find the local maximum value of ( f(x) = 4x^3 + 3x^2  6x + 1 ) using the first and second derivative tests, follow these steps:
 Find the first derivative of ( f(x) ) to locate critical points.
 Set the first derivative equal to zero and solve for ( x ) to find critical points.
 Find the second derivative of ( f(x) ) to determine concavity.
 Test the critical points found in step 2 using the second derivative test.
Let's go through these steps:

( f'(x) = 12x^2 + 6x  6 )

Setting ( f'(x) ) equal to zero and solving for ( x ): ( 12x^2 + 6x  6 = 0 ) Factor out 6: ( 6(2x^2 + x  1) = 0 ) Solve the quadratic equation ( 2x^2 + x  1 = 0 ) using the quadratic formula or factoring. The solutions are ( x = 1 ) and ( x = \frac{1}{2} ).

( f''(x) = 24x + 6 )

Test the critical points:
 For ( x = 1 ): ( f''(1) = 24(1) + 6 = 18 < 0 ). This indicates concave down, so ( x = 1 ) is a local maximum.
 For ( x = \frac{1}{2} ): ( f''\left(\frac{1}{2}\right) = 24\left(\frac{1}{2}\right) + 6 = 18 > 0 ). This indicates concave up, so ( x = \frac{1}{2} ) is a local minimum.
Therefore, the local maximum value of ( f(x) ) occurs at ( x = 1 ). To find the corresponding ( y )coordinate, plug ( x = 1 ) into ( f(x) ): [ f(1) = 4(1)^3 + 3(1)^2  6(1) + 1 = 4 + 3 + 6 + 1 = 6 ]
So, the local maximum value of ( f(x) ) is ( 6 ) when ( x = 1 ).
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