Question 29 of 35advanced🔧 ApplyShort Answer4 marks

Let f={(x,x21+x2):x∈R}f = \left\{ \left( x, \frac{x^2}{1 + x^2} \right) : x \in \mathbb{R} \right\} be a function from R\mathbb{R} into R\mathbb{R}. Determine the range of ff.

Correct Answer

The range of ff is [0,1)[0, 1).

Given f(x)=x21+x2f(x) = \frac{x^2}{1+x^2} where x∈Rx \in \mathbb{R}. The range is the set of images.

Since x2≥0x^2 \geq 0 for all real xx, and 1+x2>01 + x^2 > 0 for all real xx, we have f(x)≥0f(x) \geq 0 for all x∈Rx \in \mathbb{R}.

Also, since x2<1+x2x^2 < 1 + x^2 for all x∈Rx \in \mathbb{R}, we get f(x)=x21+x2<1f(x) = \frac{x^2}{1+x^2} < 1 for all x∈Rx \in \mathbb{R}.

At x=0x = 0: f(0)=01=0f(0) = \frac{0}{1} = 0

As ∣x∣→∞|x| \to \infty: f(x)=x21+x2=11+1x2→1f(x) = \frac{x^2}{1+x^2} = \frac{1}{1+\frac{1}{x^2}} \to 1

Since ff is continuous and takes all values from 00 to values approaching 11 (but never reaching 11), the range is [0,1)[0, 1).

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 6 | (Page 17)
For More Understanding

Explanation

The textbook states that 'The range of the function is the set of images.' Following this principle, we find all possible values of f(x)=x21+x2f(x) = \frac{x^2}{1+x^2}. The function is bounded below by 00 (attained at x=0x=0) and bounded above by 11 (approached but never attained). Since the function is continuous on R\mathbb{R}, it attains all intermediate values, giving range [0,1)[0, 1).

Solution Steps

  1. Step 1: Identify that range is the set of images {x21+x2:x∈R}\left\{\frac{x^2}{1+x^2} : x \in \mathbb{R}\right\}

  2. Step 2: Note that x2≥0x^2 \geq 0 and 1+x2>01+x^2 > 0 for all xx, so f(x)≥0f(x) \geq 0

  3. Step 3: Since x2<1+x2x^2 < 1+x^2 always, we have f(x)<1f(x) < 1

  4. Step 4: At x=0x = 0, f(0)=0f(0) = 0; as ∣x∣→∞|x| \to \infty, f(x)→1f(x) \to 1

  5. Step 5: Conclude range is [0,1)[0, 1)

PreviousNext