# Answer in Calculus for yummy #205438

3) The mean value theorem for differentiation states that if f(x) is continuous on a closed

interval [a,b] and differentiable on the open interval (a,b), then there exist a number c

in (a,b) such that

f’c=f(b)-f(a)/b-a

If f (x) =5x^2+ 3x 2 , find the value of c in the interval (2,4) that satisfies the above

theorem.

4)The mean value theorem for Differentiation states that if f(x) is differentiable on (a, b)and

continuous on [a, b], then there is at least one point c in (a, b) where f'(c)=f(b)-f(a)/b-a

Find the value of c that satisfies the theorem for the function f(x) = x 6 + in the interval [-2,

10].

“3)f(x)=5x^2+3x^2”

“f(c)=frac{f(b)-f(a)}{b-a}” where “a=2,b=4”

“f'(c)= frac{f(4)-f(2)}{4-4}=frac{f(4)-f(2)}{2}”

“f(4)=5*4^2+3*4^2=128”

“f(2)=5*2^2+3*2^2=32”

“f(c)=c-(5c^2+3c^2)”

“f'(c)= frac{d}{dc}(c-5c^2-3c^2)”

“=frac{dc}{dc}-frac{d5c^2}{dc}-frac{d3c^2}{dc}”

“=1-10c-6c”

“1-10c-6c=frac{128-32}{2}”

“1-16c=48”

“c=frac{48-1}{-16}= space -2.9375”

4) “f(x)=x^6”

“f'(c)=frac{f(b)-f(a)}{b-a}”

Where “a=-2”

“b=10”

“f'(c)=frac{f(10)-f(-2)}{10-2}”

“f(10)=10^6”

“f(-2)=-2^6”

“f'(c)=frac{d}{dc}(c-c^6)”

“=frac{dc}{dc}- frac{dc^6}{dc}”

“=1-6c^5”

“1-6c^5=frac{10^{+6}+2^6}{12}”

“c=-6.73810”