Find the absolute maximum and absolute minimum values of f on the given interval.

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

A) f(x)=2x^3-6x^2-18x+9, [-2,4]

B) f(x)=(x^2-1)^3, [-1,2]

C) f(t)= t(sqrt9-t^2), [-1,3]

Answer 1

A) f(x) = 2x³ – 6x² –18x + 9, [–2,4]

f'(x) = 6x² – 12x – 18 = 6(x² – 2x – 3) = 6(x + 1)(x – 3)

f'(x) = 0 at x = –1 and x = 3. These are x-coordinates of the critical points.

Absolute Maximum and Absolute Minimum

To find absolute maximum and absolute minimum, we evaluate f(x) at the critical points and the endpoints of the interval.

f(–2) = 5, f(–1) = 19, f(3) = –45, f(4) = –31

Absolute Maximum: f(–1) = 19

Absolute Minimum: f(3) = –45

B) f(x)=(x² – 1)³, [–1,2]

f'(x) = 3(x² – 1)² × 2x = 6x(x² – 1)² = 6x(x + 1)²(x – 1)²

The x-coordinates of the critical points are x = –1, x = 0, and x = 1

Absolute Maximum and Absolute Minimum

f(–1) = 0, f(0) = –1, f(1) = 0, f(2) = 27

Absolute Maximum: f(0) = –1

Absolute Minimum: f(2) = 27

C) f(t) = t√(9 – t²), [–1,3]

f'(t) = √(9 – t²) – t²/(√(9 – t²)) = (9 – 2t²)/(√(9 – t²))

The t-coordinates of the critical points are t = ±3/√2 ≈ ±2.12. Only +2.12 is on [–1,3]

Absolute Maximum and Absolute Minimum

f(–1) = –2√2, f(0) = –1, f(3/√2) = 9/2, f(3) = 0

Absolute Maximum: f(3/√2) = 9/2

Absolute Minimum: f(–1) = –2√2

Answer 2

A) f'(x) = 6x² - 12x - 18 = 0; x = - 1 or x = 3 f(-2) = 5, f(-1) = 19 abs max, f(3) = - 45 abs min; f(4) = -31

B) f'(x) = 6x(x² - 1)² = 0; x = 0; x = - 1, x = 1; f(-1) = 0, f(0) = -1 abs min, f(1) = 0; f(2) = 27 abs max

C) f'(t) = √9 - t² - t²/√t² - 9 = 0; 9 - 2t² = 0; t = ±3/√2; f(-1) = - 2√2 abs min, f(3/√2) = 4.5 abs max; f(3) = 0

Answer 3

First, you should find all the critical points on the given interval. For that, find x such that f’(x)=0 or does not exist. Then compare values of the function in the critical points and at the ends of the interval. The largest value is the absolute maximum, the smallest – absolute minimum.

A)   f'(x)=6x^2 -12x-18 - exists for all x.

f'(x)= 0 6x²-12x-18=0

x²-2x-3=0

x=-1, x=3 - both points belong to [-2, 4]

Now, let's plug in critical points and end points of your interval into f(x):

f(-2)=5

f(-1)=19

f(3)= - 45

f(4)= -31

Absolute maximum is 19, at x=-1. Absolute minimum is -45 at x=3.

B) f'(x)= 3(x²-1)²(2x)= 6x(x²-1)² exists for all x.

f'(x)=0 at x=-1, x=0, x=1

Plugging in those values and x=2 for the end of the interval, we get absolute min -1, absolute max 27.

C) f'=sqrt(9-t²)+t (-2t / 2sqrt(9-t²) (use product rule and chain rule for the sqrt)

Simplified: f'= (9-2t²)/sqrt(9-t²)

f'(x) DNE at sqrt(9-t²)=0, t=-3 and 3 Notice t=-3 is not on the given interval

f'=0 at x=+- sqrt (4.5) Notice t=-sqrt(4.5) is not on the given interval

Compare the values of the original f(x)

f(-1)= -2.8284 , f(sqrt(4.5))=4.5, f(3)=0

Absolute min -2.8284, absolute max 4.5.