Let f be the function given by f(x)=5cos2(x2)+ln(x+1)−3. The derivative of f is given by f′(x)=−5cos(x2)sin(x2)+1x+1. What value of c satisfies the conclusion of the Mean Value Theorem applied to f on the interval [1,4] ?

The value of c that satisfies the conclusion of the Mean Value Theorem applied to f on the interval [1,4] is approximately 0.1793.

Step-by-step explanation:

Given f(x) = 5cos²x² + ln(x + 1) - 3

The Mean Value Theorem applied to f on an interval [a, b] is given as

f'(x) = c = [f(b) - f(a)]/(b - a)

= [f(4) - f(1)]/(4 - 1)

c = [f(4) - f(1)]/3

f(4) = 5(cos4²)² + ln(5) - 3

= 3.2296

f(1) = 5(cos1²)² + ln(2) - 3

= 2.6916.

c = (3.2296 - 2.6916)/3

= 0.538/3

≈ 0.1793

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

answer: b.48 for

step-by-step explanation: