f(x)=4x^3+6x^2-24x+6

a. Find the interval on which *f* is increasing and decreasing.

b. Find the local maximum and local minimum value of *f*

c. Find the inflection point

d.Find the interval on which *f* is concave up and down

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a) You need to perform the first derivative test , hence you should find first derivative such that:

`f'(x) = 12x^2 + 12x - 24`

You need to solve the equation f'(x) = 0 such that:

`f'(x) = 0 =gt 12x^2 + 12x - 24 = 0`

`x^2 + x - 2 = 0`

`x^2 + x - 1 - 1 = 0 =gt (x^2-1) + (x-1) = 0`

(x-1)(x+1+1) = 0 => x - 1 = 0 => x = 1

`x + 2 = 0 =gt x = -2`

You need to use the signs rule for a quadratic function, hence, between the roots, the values of function have the opposite sign to the sign of coefficient of largest power, hence f'(x) < 0 for `x in (-2,1).`

You need to remember that the values of function have like signs to the sign of coefficient of largest power, hence f'(x) > 0 for `x in (-oo,-2)` and `x in (1,oo` ).

Hence, the function increases for `x in (-oo,-2)` and `x in (1,oo)` and the function decreases for `x in (-2,1).`

b) The function reaches its maximum at x=-2 and it reaches its minimum at x=1.

c) You need to solve the equation f''(x) = 0 to find the inflection points such that:

`f''(x) = 24x + 12 =gt 24x + 12 = 0 =gt 2x + 1 = 0 =gt x = -1/2`

`f(-1/2) = -4/8 + 6/4 - 24/2 + 6`

`f(-1/2) = -1/2 + 3/2 -12 + 6`

`f(-1/2) = -5`

**Hence, the inflection point is at `(-1/2,-5).` **

d) **The function is concave up for f''(x)>0, hence `x in (-1/2,oo)` and the function is concave down for`x in (-oo,-1/2).` **

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