The domain of the function f(x) = [log_{10}(f | vedclass

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Question

(B) Given the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negat

Vedclass · Accepted Answer

$1 \le x \le 4$

Vedclass · Answer

(B) Given the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative,and the argument of the logarithm must be positive.First,we require $\log_{10}(\frac{5x - x^2}{4}) \ge 0$.This implies $\frac{5x - x^2}{4} \ge 10^0$,which simplifies to $\frac{5x - x^2}{4} \ge 1$.Multiplying by $4$,we get $5x - x^2 \ge 4$,or $x^2 - 5x + 4 \le 0$.Factoring the quadratic,we have $(x - 1)(x - 4) \le 0$.This inequality holds for $x \in [1, 4]$.Since for all $x \in [1, 4]$,the term $\frac{5x - x^2}{4}$ is positive (specifically,it ranges from $1$ to $1.5625$),the logarithm is always defined.Thus,the domain is $[1, 4]$.

The domain of the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$ is

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