Question

ROLLE'S THEORUM

Answer 1

If \(f(x)=-x^2-x+6\) then clearly our secant line is the line thru \((2,0)\) with slope \(\dfrac{4-0}{2-(-2)}=\dfrac44=1\) i.e.$$y-y_0=m(x-x_0)\\y-0=1(x-2)\\y=x-2$$

Answer 2

the MVT guarantees the existence of some point \(c\in(-2,2)\) s.t. \(f'(c)=1\) here... to actually find it let's compute \(f'\)

Answer 3

$$f'(x)=\frac{d}{dx}f(x)=\frac{d}{dx}(-x^2-x+6)\\f'(x)=-2x-1$$hence we want to solve \(f'(c)=1\) which gives$$f'(c)=1\\-2c-1=1\\-2c=2\\c=-1\in(-2,2)$$ so we've found our point for (b)

Answer 4

we know that our tangent line at \(x=c\) must pass thru \((c,f(c))\) so \((-1,6)\) as \(f(-1)=6\) with slope \(f'(c)=1\) so:$$y-y_0=m(x-x_0)\\y-6=1(x-(-1))\\y-6=x+1\\y=x+7$$so our tangent line is given by \(y=x+7\)

Answer 5

now merely plot it all: http://www.wolframalpha.com/input/?i=plot+y%3D-x%5E2-x%2B6%2C+y%3Dx%2B7%2C+y%3Dx-2

http://puu.sh/6j75Z.png

Answer 6

Shouldn't the slope of secant line through (-2,4) and (2,0) be:
(0-4)/(2-(-2)) = -4/4 = -1 ?

Answer 7

correcting the slope to read \(-1\) we have$$y-y_0=m(x-x_0)\\y-0=-1(x-2)\\y=-x+2$$for our secant line and for our tangent we found$$f'(c)=-1\\-2c-1=-1\\c=0$$so it's tangent to the point \((c,f(c))\) i.e. \((0,6)\) given \(f(0)=6\) and we get$$y-y_0=m(x-x_0)\\y-6=-1(x-0)\\y=-x+6$$