Verified answer

Take the first derivative.

f'(x) = 3x^2 - 12

Find the points where the derivative is equal to zero.

3x^2 - 12 = 0

3x^2 = 12

x^2 = 4

sqrt(x^2) = +/- sqrt(4)

x = 2 or x = -2

These are the stationary points, which are local maxima or minima.

Now check the values of f(x) at the stationary points and at the endpoints of the interval.

f(-3) = (-3)^3 - 12(-3) + 1 = -27 + 36 + 1 = 10

f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17

f(2) = 2^3 - 12(2) + 1 = 8 - 24 + 1 = -15

f(5) = 5^3 - 12(5) + 1 = 125 - 60 + 1 = 66

Any of the points are eligible to be absolute maxima and minima, so look through the list. 66 is the absolute maximum, where x = 5. -15 is the absolute minimum, where x = 2.

All the stationary points are local maxima and minima. 17 is a local maximum, where x = -2. -15 is a local minimum, in addition to being an absolute minimum, where x = 2.

f is a sum of two issues -- do what's simplee355e4dab36951a7a989d4d54d2e01c and differentiate those. do no longer ingredient something out -- that purely makes it greater complicated. y = x³ - 12x y' = (x³)' - (12x)' = 3x² - 12 = 0,4 x² = 0,4; x = ±2; y = {2³-12•2e355e4dab36951a7a989d4d54d2e01c -2³-12•-2} = {-16e355e4dab36951a7a989d4d54d2e01c sixteen} yet x=-2 isn't in the intervale355e4dab36951a7a989d4d54d2e01c so this is disqualified. x=2 is in the era. the 2d by-product tells maxes from minutes: y'' = 6x = 12; + potential a min; - potential a max So there is an area min at (xe355e4dab36951a7a989d4d54d2e01cy) = (2e355e4dab36951a7a989d4d54d2e01c -sixteen) and purely the endpoints of the era can do away with the identify of absolute max/min: x=e355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01c y=0,4 x=e355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01c y=sixteen So genuinely the min is (xe355e4dab36951a7a989d4d54d2e01cy) = (2e355e4dab36951a7a989d4d54d2e01c -sixteen) and genuinely the max is (xe355e4dab36951a7a989d4d54d2e01cy) = (e355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01c sixteen) for the era [e355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01c]

Absolute maxes and mins could be either at the critical numbers or at the endpoints. Take the derivative and set to 0: 3x^2 -12=0. so x =2 or -2. at -3 the function value is -27 +36 +1 or 10.

at -2 it is -8 +24 +1 or 17. the second derivative is 6x which at -2 would be -12, so this is a local maximum. At 2, the function value is -15, and the second derivative is positive, so this is a local minimum. st 5, the function value is 66. This is the absolute max; and -15 being the lowest at x =2,

this is the absolute min.

Find the first derivative and set it equal to zero.

Then solve for x.

3x^2 -12 = 0

x = +/- 2

replace x with -3, -2, 2, 5 in x³ - 12x +1

you will get 4 answers

I assume you know the difference between absolute and local.?

Download Graph 4.4 from www.padowan.dk for free.

On "Function I Insert function", type x^3 - 12x +1, then "OK"

On "Axes", change the range of Y edge from - 20 to 20, then "OK".

> diff(x^3-12*x+1, x) = 3*x^2 - 12

> solve(3*x^2-12)

x1 = 2, x2 = - 2

> f(x):=( x^3 - 12 x +1)

> f(2) = - 15

> f(-2) = 17

( - 2, 17) = Max

(2, - 15) = Min

f(x)= x^3 - 12x + 1

dy/dx = 3x^2 - 12

3x^2 - 12 = 0

x = 2 , x = -2

f''(x) = 6x

6x = 0

6(2) = 12

So, there's a local minimum at x = 2

6(-2) = -12

and a local maximum at x = -2

Maximum - x = -2

Minimum - x = 2