Suppose f left-parenthesis x right-parenthesis equals StartFraction 1 Over 4 EndFraction x Superscript 4. Estimate f Superscript prime Baseline left-parenthesis 2 right-parenthesis, f Superscript prime Baseline left-parenthesis 3 right-parenthesis, and f Superscript prime Baseline left-parenthesis 4 right-parenthesis. What do you notice? Guess a formula for f Superscript prime Baseline left-parenthesis x right-parenthesis

f'(x) = x/2

Step-by-step explanation:

Given the function

f(x) = (1/4)x²

f'(x) will give the first derivative of the function

Using the differentiation formula

Generally, if f(y) = ayⁿ

f'(y) = nay^n-1

Applying this to differentiate the given function,

f'(x) = 2(1/4)x^2-1

f'(x) = (1/2)x

When x = 2

f'(2) = (1/2)×2

f'(2) = 1

When x = 3

f'(3) = (1/2)×3

f'(3) = 3/2

When x = 4

f'(4) = (1/2)×4

f'(4) = 2

Based on the answers, it can be seen that the values keeps increasing arithmetically. The values 1, 1 1/2, and 2 are in arithmetic progression.

The formula for calculating the nth term Tn of an arithmetic progression is expressed as:

Tn = a+(n-1)d

a is the first term of the sequence

n is the number of the terms,

d is the common difference

According to the sequence

a = 1/2, d = 3/2 - 1 = 2 - 3/2 = 1/2

Note that we started with when x=2, the first term will be at when x = 1 which will give 1/2 hence the reason for a = 1/2 instead of 1

Substituting the values in the formula

Tn = 1/2+(n-1)1/2

Tn = 1/2+n/2-1/2

Tn = n/2

Therefore the formula for f'(x) can be expressed as x/2

f'(x) = x/2

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step-by-step explanation: