Let <em>f(x)</em> = (<em>x</em> ² - 1)³. Find the critical points of <em>f</em> in the interval [-1, 2]:
<em>f '(x)</em> = 3 (<em>x</em> ² - 1)² (2<em>x</em>) = 6<em>x</em> (<em>x </em>² - 1)² = 0
6<em>x</em> = 0 <u>or</u> (<em>x</em> ² - 1)² = 0
<em>x</em> = 0 <u>or</u> <em>x</em> ² = 1
<em>x</em> = 0 <u>or</u> <em>x</em> = 1 <u>or</u> <em>x</em> = -1
Check the value of <em>f</em> at each of these critical points, as well as the endpoints of the given domain:
<em>f</em> (-1) = 0
<em>f</em> (0) = -1
<em>f</em> (1) = 0
<em>f</em> (2) = 27
So max{<em>f(x)</em> | -1 ≤ <em>x </em>≤ 2} = 27.
Answer:
its no
Step-by-step explanation:
Answer:
0.623, 6.023, 6.23, 62.30
Step-by-step explanation:
0.070 is rounded to 2 significant figures
