Over the interval, 1 < c < 4 there are two solutions of c = 1.0525 ; 2.217
Step-by-step explanation:
Given :
[tex]f(x) = 5 cos^2(x^2)+ln(x+1)-3[/tex] ----- (1)
[tex]f'(x)=-5cos(x^2)sin(x^2)+\dfrac{1}{x+1}[/tex] ------ (2)
Calculation :
According to mean value theorem,
[tex]f'(c)= \dfrac{f(b)-f(a)}{b-a}\\[/tex]
[tex]f'(c)=\dfrac{f(4)-f(1)}{4-1}=\dfrac{f(4)-f(1)}{3}\\[/tex] ------- (3)
[tex]f(4) = 5cos^2(4^2)+ln(4+1)-3\\[/tex]
[tex]f(4) = 3.2296[/tex]
[tex]f(1)= 5cos^2(1^2)+ln(1+1)-3[/tex]
[tex]f(1)=2.6916[/tex]
Now, from equation (2) and (3) we get
[tex]-5 cos(c^2)sin(c^2)+\dfrac{1}{c+1}= \dfrac {3.2269-2.6916}{3}[/tex]
[tex]-5 cos(c^2)sin(c^2)+\dfrac{1}{c+1}=0.17844[/tex]
[tex]-5sin(c)+\dfrac{2}{(c+1)} =0.35688[/tex]
Over the interval 1 < c < 4 there are two solutions of c = 1.0525 ; 2.217
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