Question

Suppose the function \(f\) is defined on the interval \([0,1]\). Find the domain of \(h\) if (a) \(h(x)=f(2 x+3)\) and (b) \(h(x)=f\left(2 x^{2}\right)\).

Short answer

The domain of \(h(x) = f(2x+3)\) is \(-\frac{3}{2} \leq x \leq -1\), and the domain of \(h(x) = f(2x^2)\) is \(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\).

Step-by-step solution

(a) Find the domain of h(x) = f(2x+3)

We are given that function \(f\) is defined on the interval \([0,1]\), which means that the input values of \(f\) lie between 0 and 1. For the function \(h(x) = f(2x+3)\), we want to find the values of x such that the input values of \(f\) lie within the given interval. That is, we want to find the domain of \(h(x)\) such that: \[ 0 \leq 2x + 3 \leq 1 \] We can now solve this inequality for x: Step 1: Subtract 3 from all parts of the inequality: \[ -3 \leq 2x \leq -2 \] Step 2: Divide by 2: \[ -\frac{3}{2} \leq x \leq -1 \] Therefore, the domain of \(h(x) = f(2x+3)\) is: \[ -\frac{3}{2} \leq x \leq -1 \]

(b) Find the domain of h(x) = f(2x^2)

Similarly, for the function \(h(x) = f(2x^2)\), we want to find the domain of \(h(x)\) such that the input values of \(f\) lie within the given interval \([0,1]\). In other words, we want to find the values of x for which: \[ 0 \leq 2x^2 \leq 1 \] We can now solve this inequality for x: Step 1: Divide by 2: \[ 0 \leq x^2 \leq \frac{1}{2} \] Step 2: Take square roots of all parts of the inequality: \[ 0 \leq x \leq \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] However, don't forget to account for the negative values of x since \(x^2\) is always non-negative. Given the inequality, we can also say: \[ 0 \geq x \geq -\frac{1}{\sqrt{2}} \] Therefore, combining both the positive and negative values, the domain of \(h(x) = f(2x^2)\) is: \[ -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \]