Question

If \[ f(x)=xe^{x^{2}-2x-3} \] is a real valued function, then \(f\) is

• A. a monotonically decreasing function
• B. an increasing function in the interval \((-1,3)\) and decreasing function in the interval \((-\infty,-1)\cup(3,\infty)\)
• C. a monotonically increasing function
• D. a decreasing function in the interval \((-1,3)\) and increasing function in the interval \((-\infty,-1)\cup(3,\infty)\)

Hint:

Whenever \[ f'(x) = (\text{positive function})\times(\text{quadratic}), \] first check whether the quadratic is always positive by completing the square. If \[ f'(x)>0 \] for all \(x\), then the function is monotonically increasing.

Answer 1

The Correct Option is C

Step 1: Differentiate the function. The given function is \[ f(x)=xe^{x^{2}-2x-3}. \] Using the product rule, \[ f'(x) = e^{x^{2}-2x-3} + xe^{x^{2}-2x-3}(2x-2). \] Taking the common factor, \[ f'(x) = e^{x^{2}-2x-3} \left[1+2x(x-1)\right]. \] Hence, \[ f'(x) = e^{x^{2}-2x-3} (2x^{2}-2x+1). \]

Step 2: Determine the sign of \(f'(x)\). Since \[ e^{x^{2}-2x-3}>0 \] for every real \(x\), the sign of \(f'(x)\) depends only on \[ 2x^{2}-2x+1. \] Now, \[ 2x^{2}-2x+1 = 2\left(x-\frac12\right)^2+\frac12. \] Clearly, \[ 2\left(x-\frac12\right)^2+\frac12>0 \] for every real number \(x\). Therefore, \[ f'(x)>0 \] for all \(x\in\mathbb{R}\).

Step 3: Conclude the monotonicity. Since \[ f'(x)>0 \] throughout its domain, \[ \boxed{f(x)\text{ is monotonically increasing on }\mathbb{R}.} \] Hence, the correct option is \(\boxed{(C)}\).