For small values of h, the function (16+h)^1/4 is best approximately by which of the following?
a. 4+h/32
b. 2+h/32
c. h/32
d. 4-h/32
e. 2-h/32
Even without doing much figuring, you know that it is near 2. That reduces your choices to B and E.
Since you have 16+h, it will naturally be B.
Now, the binomial theorem says that (16+h)^1/4 = 16^1/4 + 1/4 * 16^-3/4 * h + ...
= 2 + 1/4 * 1/8 * h
= 2 + h/32
Well, I must say, these options are giving me some mathematical jokes!
Let's have some fun with this question.
For small values of h, we can approximate (16+h)^1/4 by using a technique called the binomial theorem. This theorem allows us to expand expressions involving powers of sums or differences, in this case, (16+h).
So let's apply some clown magic here!
Using the binomial theorem, we can approximate (16+h)^1/4 as:
16^1/4 + (1/4)(16^(-3/4))(h) + ...
But wait a minute, let's simplify this:
2 + (1/16)(h) + ...
And look, we have an answer that matches one of the options!
So the best approximation for (16+h)^1/4 for small values of h is:
b. 2+h/32
I hope you enjoyed your clown-ful answer!
To find the best approximation for the function (16+h)^(1/4) for small values of h, we can use the binomial approximation formula.
The binomial approximation formula states that for small values of h, (a+b)^n is approximately equal to a^n + n*a^(n-1)*b.
In our case, a = 16, b = h, and n = 1/4.
Using the formula, we can approximate (16+h)^(1/4) as:
= 16^(1/4) + (1/4)*16^(-3/4)*h
Simplifying this expression, we get:
= 2 + (1/4)*2^(-3)*h
= 2 + (1/32)*h
= 2 + h/32
Therefore, the best approximation for the given function for small values of h is given by option b. 2+h/32.
To find the best approximation for the function (16+h)^(1/4) for small values of h, we can use the Taylor series expansion.
The Taylor series expansion of a function f(x) centered at a point a is given by:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
In this case, we have f(x) = (16+x)^(1/4) and we want to find the best approximation for small values of x. We can center the expansion at x = 0, so a = 0.
First, let's find the derivatives of f(x):
f'(x) = (1/4)(16+x)^(-3/4)
f''(x) = (-3/16)(16+x)^(-7/4)
f'''(x) = (21/64)(16+x)^(-11/4)
...
Now, let's evaluate the function and its derivatives at x = 0:
f(0) = (16+0)^(1/4) = 2
f'(0) = (1/4)(16+0)^(-3/4) = 1/8
f''(0) = (-3/16)(16+0)^(-7/4) = -3/128
f'''(0) = (21/64)(16+0)^(-11/4) = 21/4096
...
Now we can plug these values into the Taylor series expansion and simplify to find the best approximation for small values of x:
f(x) ≈ f(0) + f'(0)x + f''(0)(x^2)/2! + f'''(0)(x^3)/3! + ...
f(x) ≈ 2 + (1/8)x + (-3/128)(x^2)/2 + (21/4096)(x^3)/6 + ...
Simplifying further:
f(x) ≈ 2 + (1/8)x - (3/256)x^2 + (7/2048)x^3 + ...
Comparing this expression with the given options:
a. 4+h/32 → incorrect, doesn't match the pattern of the Taylor series expansion
b. 2+h/32 → correct, matches the pattern of the Taylor series expansion
c. h/32 → incorrect, doesn't match the pattern of the Taylor series expansion
d. 4-h/32 → incorrect, doesn't match the pattern of the Taylor series expansion
e. 2-h/32 → incorrect, doesn't match the pattern of the Taylor series expansion
Therefore, the best approximation for (16+h)^(1/4) for small values of h is given by option b. 2+h/32.