How do you find the critical points for $- \frac{2}{9} x^{3} - \frac{2}{3} x^{2} + \frac{16}{3} x + \frac{160}{9}$ $-2/9x^3 -2/3x^2+16/3x+160/9$ and the local max and min?

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Answer 1 · 2016-09-28T14:14:47

See below.

$f (x) = - \frac{2}{9} x^{3} - \frac{2}{3} x^{2} + \frac{16}{3} x + \frac{160}{9}$ $f(x) = -2/9x^3 -2/3x^2+16/3x+160/9$

First, let's clean the expression up by factoring out $- \frac{2}{9}$ $-2/9$ .

$f (x) = - \frac{2}{9} (x^{3} + 3 x^{2} - 24 x - 80)$ $f(x) = -2/9(x^3 + 3x^2 - 24x - 80)$ .

Now, we'll find the critical numbers by finding the numbers in the domain at which the derivative is $0$ $0$ or fails to exist.

The domain of $f$ $f$ is all real numbers.

$f'(x) = -2/9(3x^2+6x-24)$ .

The derivative exists for all real $x$ , so now we need to solve $f'(x)=0$ to find the critical numbers.

$-2/9(3x^2+6x-24)=0$ is equivalent to

$-2/3(x^2+2x-8)=0$

$-2/3(x+4)(x-2) = 0$

$x=-4$ $" "$ or $" "$ $x=2$ .

These are the critical numbers for $f$ . They determine three intervals on the number line.

We look at the sign of $f'$ on each interval to determine whether $f$ is increasing or decreasing on the interval

${: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"), ((-oo,-4)," " -" ", " "" Decr"), ((-4,2), " " +, " " " Incr"), ((2 ,oo), " " -, " "" Decr") :}$

$f$ has a local minimum of $0$ at $-4$ , and

$f$ has a local maximum of 24 at $2$ .

Calculations

$f(-4) = -2/9[(-4)^3+3(-4)^2-24(-4)-80]$

$= -2/9[-4(16)+3(16)+6(16)-5(16)]$

$= -2/9[0(16)] = 0$

and

$f(2) = -2/9[(2)^3+3(2)^2-24(2)-80]$

$= -2/9[2(4)+3(4)-12(4)-20(4)]$

$= -2/9[-27(4)] = (2cancel((27))^3(4))/cancel(9) = 24$

How do you find the critical points for -2/9x^3 -2/3x^2+16/3x+160/9−29x3−23x2+163x+1609-2/9x^3 -2/3x^2+16/3x+160/9 and the local max and min?