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Use the First Derivative Test to find the points of local maxima and minima of the function $ƒ(x)=4x^3+3x^2−6x+1$.

The final answer is expected to be in the form of a table containing all the required information about f(x). Once the table is created, add the property row “Increase/decrease of f(x)”, add intervals, where f(x) is increasing/decreasing, and specify local maxima and minima.

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  • $\begingroup$ what have you done? $\endgroup$ Oct 29, 2014 at 20:18
  • $\begingroup$ $f'(x) = 12x^2 + 6x - 6 $ $\endgroup$
    – Csci319
    Oct 29, 2014 at 20:19

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we have $f'(x)=6(2x^2+x-1)=6(2x^2+x-1)=0$ solving this equation we obatien $(x+1)(x-1/2)=0$ solving $(x+1)(2x-1)\geq 0$ we get $-\infty<x\le -1$ or $\frac{1}{2}\le x<\infty$ in the other case $-1<x<\frac{1}{2}$ is $f(x)$ strictly monotonously decreasing

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  • $\begingroup$ so critical points are $x = -1 and x = 1/2$ then you use those to plug in for the maxima and minima? $\endgroup$
    – Csci319
    Oct 29, 2014 at 20:26
  • $\begingroup$ yes i did this and posted you the monotonicity $\endgroup$ Oct 29, 2014 at 20:29
  • $\begingroup$ to get maxima and minima $f(-1) = 4(-1)^3 +3(-1)^2 + 6 (-1) - 6 = -13$ $f(1/2) = 4(1/2)^3 +3(1/2)^2 + 6 (1/2) - 6 = -7/4$ so is the min -13 at x=-1 and max -7/4 at x= 1/2? $\endgroup$
    – Csci319
    Oct 29, 2014 at 20:35
  • $\begingroup$ is this correct? $\endgroup$
    – Csci319
    Oct 31, 2014 at 15:46

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