find the absolute maximum and absolute minimum values of the function\n\n$f(x)=x^{3}+12x^{2}-27x +…

find the absolute maximum and absolute minimum values of the function\n\n$f(x)=x^{3}+12x^{2}-27x + 9$\n\nover each of the indicated intervals.\n\n(a) the interval $=-10,0$.\n\n1. absolute maximum=\n2. absolute minimum=\n\n(b) the interval $=-7,2$.\n\n1. absolute maximum=\n2. absolute minimum=\n\n(c) the interval $=-10,2$.\n\n1. absolute maximum=\n2. absolute minimum=
Answer
Explanation:
Step1: Find the derivative of the function
The derivative of (f(x)=x^{3}+12x^{2}-27x + 9) is (f^\prime(x)=3x^{2}+24x-27 = 3(x^{2}+8x - 9)=3(x + 9)(x-1))
Step2: Find the critical points
Set (f^\prime(x)=0), then (3(x + 9)(x - 1)=0). The critical points are (x=-9) and (x = 1)
Part (a) Interval ([-10,0])
Step3: Evaluate the function at critical points and endpoints
- (f(-10)=(-10)^{3}+12(-10)^{2}-27(-10)+9=-1000 + 1200+270 + 9=479)
- (f(-9)=(-9)^{3}+12(-9)^{2}-27(-9)+9=-729+972 + 243+9=504)
- (f(0)=0^{3}+12\times0^{2}-27\times0 + 9=9)
Part (b) Interval ([-7,2])
Step4: Evaluate the function at critical points and endpoints
- (f(-7)=(-7)^{3}+12(-7)^{2}-27(-7)+9=-343+588+189 + 9=443)
- (f(1)=1^{3}+12\times1^{2}-27\times1 + 9=1+12-27 + 9=-5)
- (f(2)=2^{3}+12\times2^{2}-27\times2 + 9=8+48-54 + 9=11)
Part (c) Interval ([-10,2])
Step5: Evaluate the function at critical points and endpoints
- (f(-10)=479) (calculated in part (a))
- (f(-9)=504) (calculated in part (a))
- (f(1)=-5) (calculated in part (b))
- (f(2)=11) (calculated in part (b))
Answer:
(a)
- Absolute maximum (=504)
- Absolute minimum (=9)
(b)
- Absolute maximum (=443)
- Absolute minimum (=-5)
(c)
- Absolute maximum (=504)
- Absolute minimum (=-5)