Expert: Ahmed Salami - 9/19/2010

Question
Here is the problem.
f(x)=cosx^2      

Compute the slope of the secant line between the points at(a)x=1 and x=2  (b) x=2 and x=3  (c) x=1.5 and x=2  (d)x=2 and x=2.5  (e) x=1.9 and x=2  (f) x=2 and x=2.1   (g) use parts(a)-(f) and other calculations as needed to estimate the slope of the tangent line at x=2.

Sorry for the long problem. Thank you for your help =)  

Answer
Hi Seth,
I hope you meant cos(x²) and not cos²x.
Anyway, for f(x) = cos(x²), the slope of the secant line between two points x = a and x = b is given by
[cos(b²)-cos(a²)]/(b-a)

a) Slope = [cos(2²)-cos(1²)]/(2-1) = (-0.6536 - 0.5403)/1
= -1.1939
b) Slope = [cos(3²)-cos(2²)]/(3-2) = (-0.9111 - -0.6536)/1
= -0.2575
c) Slope = [cos(2²)-cos(1.5²)]/(2-1.5) = (-0.6536 - -0.6282)/0.5
= -0.0508
d) Slope = [cos(2.5²)-cos(2²)]/(2.5-2) = (0.9994 - -0.6536)/0.5
= 3.3060
e) Slope = [cos(2²)-cos(1.9²)]/(2-1.9) = (-0.6536 - -0.8923)/0.1
= 2.3870
a) Slope = [cos(2.1²)-cos(2²)]/(2.1-2) = (-0.2978 - -0.6536)/0.1
= 3.5580

Considering pairs of x values which are (lets say, equidistant) from x = 2 as in x = 1 & 3, x = 1.5 & 2.5, x = 1.9 & 2.1. We could try to make an estimate of the slope at x = 2 by taking the average of the slopes of the secant lines at pair of values.
For x = 1 & 3, estimated slope = (-1.1939 + -0.2575)/2 = -0.7257
For x = 1.5 & 2.5, estimated slope = (-0.0508 + 3.3060)/2 = 1.6276
For x = 1.9 & 2.1, estimated slope = (2.3870 + 3.5580)/2 = 2.9725

These slope values differ considerably but we can see that as the pair of points move closer to x = 2 the estimated slope approaches, say, a value of 3.

Analytically, the slope function of f(x) = cos(x²) is given by
f'(x) = -2x.sin(x²)
and at x = 2
f'(2) = -2(2).sin(4) = -4(-0.7568) = 3.0272

Note that x is in radians and not degrees!!!

Regards.