Start with #47. To find the critical values, you must differentiate this function. x times (4-x)^3 is a product, so use the product rule. The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x]
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of f '(x) = (4-x)^2 [-3x+4-x]. (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1. Thus, the "cv" are {4,1}.
Evaluate the given function at x: {4,1}. For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8. Thus, one of the extreme values is (1,8).
Answer:
1 2/10
Step-by-step explanation:
Or it could be 6/5 or 1.2
Answer:
900 push-ups
Step-by-step explanation:
75 push-ups in one week
In 12 weeks = 75 × 12 = 900 push-ups
<em>PLEASE DO</em><em> </em><em>MARK ME</em><em> </em><em>AS BRAINLIEST</em><em> </em><em>IF MY</em><em> </em><em>ANSWER IS</em><em> </em><em>HELPFUL</em><em> </em><em>:</em><em>)</em><em> </em>
E)6,2 is la reposta espero te sirva Friends
Step-by-step explanation:
tanks por los points
