If f(x) is a real function defined on [-1, 1] | vedclass

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(B) Given that the function $f(x)$ is defined on the interval $[-1, 1]$.For the function $g(x) = f(5x + 4)$ to be defined, the argument $(5x + 4)$ mus

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$[-1, -\frac{3}{5}]$

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(B) Given that the function $f(x)$ is defined on the interval $[-1, 1]$.For the function $g(x) = f(5x + 4)$ to be defined, the argument $(5x + 4)$ must lie within the domain of $f$, which is $[-1, 1]$.Therefore, we have the inequality:$-1 \leq 5x + 4 \leq 1$Subtracting $4$ from all parts of the inequality:$-1 - 4 \leq 5x \leq 1 - 4$$-5 \leq 5x \leq -3$Dividing all parts by $5$:$-1 \leq x \leq -\frac{3}{5}$Thus, the function $g(x)$ is defined on the interval $[-1, -\frac{3}{5}]$.

If $f(x)$ is a real function defined on $[-1, 1]$, then the function $g(x) = f(5x + 4)$ is defined on the interval

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