Let f be the function given by f(x) = x+4(x−1)(x+3) on the closed interval [−5,5]. On which closed interval is the function f gu
aranteed by the Extreme Value Theorem to have an absolute maximum and an absolute minimum?
1 answer:
Answer:
Step-by-step explanation:
f(x)=x+4(x^2+2x-3)=4x^2+9x-12
f'(x)=8x+9
f'(x)=0,gives x=-9/8
f(-5)=-5+4(-5-1)(-5+3)=-5+4*-6*-2=43
f(-9/8)=-9/8+4(-9/8-1)(-9/8+3)
=-9/8+4*-17/8*15/8
=-9/8-255/16
=-273/16=-17 1/16
f(5)=4*5^2+9*5-12=100+45-12=133
absolute maximum=133
absolute minimum=-17 1/16
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