Asked by kale
limit as delta x approaches 0 when f(x) = 4/ sqr x
Have to use the (F(x+delta x)-f(x))/delta x method
Have to use the (F(x+delta x)-f(x))/delta x method
Answers
Answered by
Steve
just plug and chug
[f(x+h)-f(x)]/h
4(x+h)^(-1/2) - 4*x^(-1/2)
Now, to evaluate (x+h)^(-1/2), use the binomial theorem:
(a+b)^n = a^n + n*a^(n-1)*b^1 + n(n-1)/2 * a^(n-1)b^2 + ...
(x+h)^(-1/2) = x^(-1/2) + (-1/2)*x^(-3/2)*h + (-1/2)(-3/2)/2 * x^(-5/2)h^2 + ...
(x+h)^(-1/2) - x^(-1/2) = (-1/2)*x^(-3/2)*h + (-1/2)(-3/2)/2 * x^(-5/2)h^2 + ...
divide all that by h to get
(-1/2)*X^(-3/2) + (-1/2)(-3/2)/2 * x^(-5/2)h + ...
= -1/2x^(-3/2) + 3/8x^(-5/2)*h + ...
As h->0, all the terms with h disappear, leaving just
-1/2 x^(-3/2)
multiply that by 4 from the original function, and you get
-2x^-3/2 = -2/√x<sup>3</sup>
[f(x+h)-f(x)]/h
4(x+h)^(-1/2) - 4*x^(-1/2)
Now, to evaluate (x+h)^(-1/2), use the binomial theorem:
(a+b)^n = a^n + n*a^(n-1)*b^1 + n(n-1)/2 * a^(n-1)b^2 + ...
(x+h)^(-1/2) = x^(-1/2) + (-1/2)*x^(-3/2)*h + (-1/2)(-3/2)/2 * x^(-5/2)h^2 + ...
(x+h)^(-1/2) - x^(-1/2) = (-1/2)*x^(-3/2)*h + (-1/2)(-3/2)/2 * x^(-5/2)h^2 + ...
divide all that by h to get
(-1/2)*X^(-3/2) + (-1/2)(-3/2)/2 * x^(-5/2)h + ...
= -1/2x^(-3/2) + 3/8x^(-5/2)*h + ...
As h->0, all the terms with h disappear, leaving just
-1/2 x^(-3/2)
multiply that by 4 from the original function, and you get
-2x^-3/2 = -2/√x<sup>3</sup>
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