Question

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].

Answer 1

$3)f(x)=5x^2+3x^2$

$f(c)=\frac{f(b)-f(a)}{b-a}$ 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 $a=-2$

$b=10$

$f'(c)=\frac{f(10)-f(-2)}{10-2}$

$f(10)=10^6$

$f(-2)=-2^6$

$f'(c)=\frac{d}{dc}(c-c^6)$

$=\frac{dc}{dc}- \frac{dc^6}{dc}$

$=1-6c^5$

$1-6c^5=\frac{10^{+6}+2^6}{12}$

$c=-6.73810$