# Finding local Maxima and minima then create a table

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.

• what have you done? – Dr. Sonnhard Graubner Oct 29 '14 at 20:18
• $f'(x) = 12x^2 + 6x - 6$ – Csci319 Oct 29 '14 at 20:19

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
• so critical points are $x = -1 and x = 1/2$ then you use those to plug in for the maxima and minima? – Csci319 Oct 29 '14 at 20:26
• 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? – Csci319 Oct 29 '14 at 20:35