Chapter 2: Differentiation
Section 2.1: What Is a Derivative?
Example 2.1.2
Show that at x=2 a unique slope cannot be assigned to the graph of fx=x2−4+9. Consequently, a tangent line does not exist at 2,3 on this curve.
Solution
In Figure 2.1.2(a), the green dot on the graph of fx is at the point 2,3 where no unique slope can be assigned.
To see this, consider secant lines through 2,3 and 2+h,f2+h, the slopes of which are
f2+h−32+h−2
The limiting values for these slopes are then±2/3, the positive value when the limit is taken from the right; the negative, from the left.
Consequently, no limit exists for these slopes, and therefore a unique slope cannot be assigned to the graph of f at x=2.
p1:=plot(sqrt(abs(x^2-4)+9),x=-3..3): p2:=plot([[2,3]],style=point,symbol=solidcircle,symbolsize=15,color=green): plots:-display(p1,p2,labels=[x,y],view=[-3..3,0..5],scaling=constrained);
Figure 2.1.2(a) Graph of fx=x2−4+9
The "corner" at x=2 on the graph of f is an example of a point where a tangent line does not exist. The following details demonstrate this computationally.
Define the function f
Control-drag fx=… Context Panel: Evaluate and Display Inline
fx=x2−4+9→assign as functionf
Limiting slopes of the secant lines
Expression palette: Limit operator Limit from the left
limh→0−f2+h−3h = −23
Expression palette: Limit operator Limit from the right
limh→0+f2+h−3h = 23
Stepwise computation of the limits:
=2+h2−4+9−32+h−2
=h2+4 h+9−3h
=h2+4 h+9−3h ⋅ h2+4 h+9+3h2+4 h+9+3
=h2+4 hh (h2+4 h+9+3)
=hh h+4|h2+4 h|+9+3
When computing the limit from the right, h>0, so h/h=1; but from the left, h<0, so h/h=−1. The fraction
h+4|h2+4 h|+9+3
tends to 4/6=2/3 as h→0, so the limit from the right is 2/3 but the limit from the left is −2/3.
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