SOLUTION: Please help me with this problem: Suppose f(x) = {{{x^2 + 2}}}. Find the slope of the line which passes through the points(0,f(0)) and (2,f(2)). Illustrate the mean value theorem

Algebra ->  Test -> SOLUTION: Please help me with this problem: Suppose f(x) = {{{x^2 + 2}}}. Find the slope of the line which passes through the points(0,f(0)) and (2,f(2)). Illustrate the mean value theorem      Log On


   



Question 1032916: Please help me with this problem:
Suppose f(x) = x%5E2+%2B+2. Find the slope of the line which passes through the points(0,f(0)) and (2,f(2)). Illustrate the mean value theorem for derivatives by finding a number c in (0,2) such that f'(c) equals the slope of the line described.

Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The two points are (0,2) and (2,6). The slope of the line passing through them is %286-2%29%2F%282-0%29+=+4%2F2+=+2.
f'(x) = 2x.
Since f(x) is differentiable on the interval (0,2) and continuous on the interval [0,2], there is a c in the interval (0,2) such that f'(c) = %28f%282%29+-+f%280%29%29%2F%282-0%29+=+2+=+2c.
Indeed, 2c = 2 ==> c = 1.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

We draw the graph of f(x) = x²+2.

We find f(0) = 0²+2 = 2 and f(2) = 2²+2 = 6

Then the points (0,f(0)) and (2,f(2)) are (0,2) and (2,6),

We plot those and draw a line through them
(the green line below): 



Next we find the slope of that green line, using the
slope formula from algebra:

m = %28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29
where (x1,y1) = (0,2)
and where (x2,y2) = (2,6)

m = %286-2%29%2F%282-0%29 = 4%2F2 = 2

Now we want to find a point on the graph, between those
two points where the slope of a tangent line is parallel
to the green line.

The derivative is a formula for the slope of the tangent
line at any point we substitute the x-coordinate in.

f(x) = x²+2
f'(x) = 2x

We set 2x equal to the slope 2 of that green line.

2x = 2
 x = 1

That means that at the point (1,f(1)) which is (1,1²+2)
or (1,3), if we draw a line tangent to the graph at that
point it will be parallel to the green line.  So we draw
a tangent line at the point (1,3) (in blue), and we notice
that it is parallel to the green line:



I have drawn a black line from the point of tangency
of the blue line down to the x-axis to show that the value of 
c on the x-axis in the interval (0,2) is the value c=1.

What this is all about is showing you that if you have a
line that cuts a curve in two points there is always a point 
between them where you can draw a tangent line paralell to
that line that cuts through the curve.  Then the x-coordinate
of that point of tangency is the value of c.

Edwin