4.a.f(x) =4.2x2, f'(6) = 1m
ri, :8?G:i)$19 = -2.4 = x+6 x-b b. Graph of the difference quotient
15. Local Linearity Problem
a. First find f'(1), then plot a straight line through point (1, f(1)) using f'(1) as the slope. b. Near the point (1, f(1)), which is (1 , 1), the tangent line and the curve appear coincidental. c. The curve appears to get closer and closer to the line and finally touch it at the point (1, f(1)). d. Near point (1 ,1 ) the curve looks linear. e. lf a graph has local linearity, the graph near that point looks like the tangent line. Therefore, the derivative at that point cbuld be said to equal the slope of the graph at that point. 16. Local Nonlinearitv Problem
c. and d. Graph. Tangent line: y = -2.4x + 7.2
i-Z - 1)f(1) = 12 + 0.1(1 - 1)2t3 - 1 + 0 = 1, o.E.D. f(x)=
b. Zoom by a factor of 10,000. Graph
tim LL12)4i3) =x+-2 X+Z
ri, lLil)Gj2) =x+-4 X+4
tim fU)Idi-Ql:2)=4 = x+1 X-l x3-x2:-4x+6-8 8. f(-1) = x+1 ,(E11
c. The graph has a cusp at x = 1. lt changes direction abruptly, not smoothly. d. lf you draw a secant line through (1 ,1) from a point just to the left of x = 1, it has a large negative slope. lf you draw one from a point just to the right, it has a large positive slope. ln both cases, the secant line becomes vertical as x approaches 1, and a vertical line has infinite slope. So there is no real number equal to the derivative.
e.f(3)=, -ozlits -0.7(x-3) _ __r\7 lim = x+3 x-5
.. 1.3x-3-2.2 rrm )(_+
10. f'(4) =
rim 1j0i0 = xs4 X-4
11.f(-1)= 12. f'(3)
13. The derivative of a linear function equals the slope.
The tangent line coincides with the graph. 14. The derivative of a constant function is zero. Constant functions don't change! The tangent line coincides with the graph.
Published on Sep 17, 2010
12. f'(3) = lqt ta.= =O x+-2 X+Z _ i-Z a. f(x)= x2 + 0'1(!x - 1)- f(1) = 12 + 0.1(1 - 1)2t3 - 1 + 0 = 1, o.E.D. x+-4 X+4 Solufions Mqnuol 27...