Saturday, November 30, 2013

Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 48

a.) Suppose that g(x)=x23, show that g(0) does not exist.

Using the definition of derivative


g(x)=limh0g(x+h)g(x)hg(x)=limh0(x+h)23(x)23hSubstitute g(x+h) and g(x)g(x)=limh0(x+h)23(x)23h(x+h)43+(x)23+(x+h)23+(x)43(x+h)43+(x)23+(x+h)23+(x)43Multiply both numerator and denominator by [(x+h)43+(x)23(x+h)23+(x)43]. Recall that (a3b3)=(ab)(a2+ab+b2)g(x)=limh0(x+h)2\cancel(x)23(x+h)43+\cancel(x)23(x+h)43\cancel(x)43(x+h)23+\cancel(x)43(x+h)23x2(h)[(x+h)43+(x)23(x+h)23+(x)43]Combine like termsg(x)=limh0(x+h)2x2(h)[(x+h)43+(x)23(x+h)23+(x)43]Expand the equationg(x)=limh0\cancelx2+2xh+h2\cancelx2(h)[(x+h)43+(x)23(x+h)23+(x)43]Combine like termsg(x)=limh02xh+h2(h)[(x+h)43+(x)23(x+h)23+(x)43]Factor the numeratorg(x)=limh0\cancelh(2x+h)\cancel(h)[(x+h)43+(x)23(x+h)23+(x)43]Cancel out like termsg(x)=limh0[2x+h(x+h)43+(x)23(x+h)23+(x)43]=2x+0(x+0)43+(x)23(x+0)23+(x)43=2x(x)43+(x)23(x)23+(x)43=2x(x)43+(x)43+(x)43Evaluate the limitg(x)=2x3(x)43Factor the denominatorg(x)=2\cancelx3\cancel(x)(x)13Cancel out like termsg(x)=23(x)13Substitute x which is zerog(0)=23(0)13Simplify the equationg(0)=20


The function does not exist because the denominator is zero.

b.) Suppose that a0, find g(a).

Using the definition of derivative


f(a)=limxaf(x)f(a)xaf(a)=limxa(x)23(a)23xaSubstitute f(x) and f(a)f(a)=limxa(x)23(a)23xa(x)43+(ax)23+(a)43(x)43+(ax)23+(a)43Multiply both numerator and denominator by [(x)43+(ax)23+(a)43]. Recall that (a3b3)=(ab)(a2+ab+b2)f(a)=limxax2\cancel(a)23(x)43+\cancel(a)23(x)43\cancel(a)43(x)23+\cancel(a)43(x)23a2(xa)[(x)43+(ax)23+(a)43]Combine like termsf(a)=limxax2a2(xa)[(x)43+(ax)23+(a)43]Factor the numeratorf(a)=limxa\cancel(xa)(x+a)\cancel(xa)[(x)43+(ax)23+(a)43]Cancel out like termsf(a)=limxa[x+a(x)43+(ax)23+(a)43]=a+a(a)43+(aa)23+(a)43=2a(a)43+(a)43+(a)43Evaluate the limitf(a)=2a3(a)43Factor the denominatorf(a)=2\cancela3\cancel(a)(a)13Cancel out like termsf(a)=23(a)13 or 233a


c.) Prove that y=x23 has a vertical tangent line at (0,0)

If the function has a vertical tangent line at x=0,limx0f(x)=

Given that, f(x)=23(x)13

Suppose that we substitute a value closer to 0 from the left and the right of the limit of f(x), let's say x=0.0000001 and x=0.0000001


limx02330.0000001=143.629limx0+2330.0000001=143.629


This means that where x gets closer and closer to from the left, the value of the limit approaches a very large negative number. On the other hand, as x gets closer and closer to from the right, the value of the limit approaches a very large positive number. The tangent lines with these values become steeper and steeper as x approaches until such time that the tangent line becomes a vertical line @ x=0.

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