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# Verify Rolle's theorem for the function $f (x) =x^2 + 2x - 8, x \: \in [ -4 , 2 ].$

Toolbox:
• Let $f:[a,b]\rightarrow R$ be continuous on [a,b] and differentiable on (a,b).Such that $f(a)=f(b)$ where a and b are some real numbers.Then there exists some $c$ in $(a,b)$ such the $f'(c)=0$
Step 1:
Now $f(x)=x^2+2x-8$ is a polynomial.
$\Rightarrow$ It is continuous and derivable in its domain $x\in R$.Hence it is continuous in the interval $[4,2]$ and derivable in the interval $[-4,2]$
$f(-4)=(-4)^2+2(-4)-8$
$\qquad\;\;=16-8-8$
$\qquad\;\;=0$
$f(2)=2^2+2(2)-8$
$\qquad\;\;=4+4-8$
$\qquad\;\;=0$
Step 2:
$f(x)=2x+2$
$\quad\;\;=2c+2=0$
$2c=-2$
$c=-1$
$f'(c)=0$ at $c=-1$
Hence the condition of Rolle's theorem is satisfied
edited May 13, 2013