Question 20 of 82beginner🔧 ApplyShort Answer3 marks

Show that the function given by f(x) = e²ˣ is increasing on R.

Correct Answer

Given the function f(x)=e2xf(x) = e^{2x}. Differentiating with respect to xx, we get f′(x)=2e2xf'(x) = 2e^{2x} (using the Chain Rule).

For any real number xx, the exponential term e2xe^{2x} is always positive. Therefore, the derivative f′(x)=2e2xf'(x) = 2e^{2x} is strictly greater than zero for all x∈Rx \in \mathbf{R}.

Since f′(x)>0f'(x) > 0 for all xx in the domain, the function ff is increasing on R\mathbf{R}.

Exercise: EXERCISE 6.2 | Q: 2 | (Page 12)
For More Understanding

Explanation

This solution follows the methodology established in Example 8 of the provided context. Just as Example 8 demonstrates that a polynomial function is increasing on R by showing its derivative is always positive, this solution applies the same principle to the exponential function. The context mentions the Chain Rule in Example 35, which justifies the differentiation step. The conclusion relies on the theorem mentioned in the Remarks section: if the derivative is positive for all x in an interval, the function is increasing.

Solution Steps

  1. Step 1: Calculate the derivative of the function f(x)=e2xf(x) = e^{2x}. Using the Chain Rule, f′(x)=2e2xf'(x) = 2e^{2x}.

  2. Step 2: Analyze the sign of the derivative. Since e2x>0e^{2x} > 0 for all x∈Rx \in \mathbf{R}, it follows that f′(x)=2e2x>0f'(x) = 2e^{2x} > 0 for all x∈Rx \in \mathbf{R}.

  3. Step 3: Conclude based on the sign. Because f′(x)>0f'(x) > 0 for every point in the domain, the function is increasing on R.

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