a.) Suppose that g(x)=x23, show that g′(0) does not exist.
Using the definition of derivative
g′(x)=limh→0g(x+h)−g(x)hg′(x)=limh→0(x+h)23−(x)23hSubstitute g(x+h) and g(x)g′(x)=limh→0(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 (a3−b3)=(a−b)(a2+ab+b2)g′(x)=limh→0(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)23−x2(h)[(x+h)43+(x)23(x+h)23+(x)43]Combine like termsg′(x)=limh→0(x+h)2−x2(h)[(x+h)43+(x)23(x+h)23+(x)43]Expand the equationg′(x)=limh→0\cancelx2+2xh+h2−\cancelx2(h)[(x+h)43+(x)23(x+h)23+(x)43]Combine like termsg′(x)=limh→02xh+h2(h)[(x+h)43+(x)23(x+h)23+(x)43]Factor the numeratorg′(x)=limh→0\cancelh(2x+h)\cancel(h)[(x+h)43+(x)23(x+h)23+(x)43]Cancel out like termsg′(x)=limh→0[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 a≠0, find g′(a).
Using the definition of derivative
f′(a)=limx→af(x)−f(a)x−af′(a)=limx→a(x)23−(a)23x−aSubstitute f(x) and f(a)f′(a)=limx→a(x)23−(a)23x−a⋅(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 (a3−b3)=(a−b)(a2+ab+b2)f′(a)=limx→ax2−\cancel(a)23(x)43+\cancel(a)23(x)43−\cancel(a)43(x)23+\cancel(a)43(x)23−a2(x−a)[(x)43+(ax)23+(a)43]Combine like termsf′(a)=limx→ax2−a2(x−a)[(x)43+(ax)23+(a)43]Factor the numeratorf′(a)=limx→a\cancel(x−a)(x+a)\cancel(x−a)[(x)43+(ax)23+(a)43]Cancel out like termsf′(a)=limx→a[x+a(x)43+(ax)23+(a)43]=a+a(a)43+(a⋅a)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 233√a
c.) Prove that y=x23 has a vertical tangent line at (0,0)
If the function has a vertical tangent line at x=0,limx→0f′(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
limx→0−233√−0.0000001=−143.629limx→0+233√0.0000001=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.
Saturday, November 30, 2013
Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 48
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