Find the Domain y = log base 4 of 4x^3-x+1

Problem

y=(log_4)(4*x3−x+1)

Solution

1. Identify the condition for the domain of a logarithmic function. The argument of the logarithm must be strictly greater than zero.

4*x3−x+1>0

2. Analyze the cubic polynomial ƒ(x)=4*x3−x+1 to find its roots. By testing small rational values or using the rational root theorem, we find no simple integer or rational roots.

3. Find the local extrema of ƒ(x) by taking the derivative.

d(4*x3−x+1)/d(x)=12*x2−1

4. Solve for the critical points where the derivative is zero.

12*x2−1=0

x2=1/12

x=±1/(2√(,3))

5. Evaluate the function at the local minimum x=1/(2√(,3)) to see if the function ever drops below zero.

ƒ(1/(2√(,3)))=4*(1/(24√(,3)))−1/(2√(,3))+1

ƒ(1/(2√(,3)))=1/(6√(,3))−3/(6√(,3))+1

ƒ(1/(2√(,3)))=1−1/(3√(,3))

6. Determine the sign of the local minimum. Since 3√(,3)=√(,27)>1 the fraction 1/(3√(,3)) is less than 1 meaning ƒ(1/(2√(,3)))>0

7. Conclude that since the local minimum is positive and the function approaches ∞ as x→∞ we only need to find the single real root where the function crosses from negative to positive. Using the cubic formula or numerical methods, the root is approximately x≈−0.657

8. Express the domain as the interval where the polynomial is positive.

Final Answer

Domain:x>−0.6573...

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