For x ∈ R, let ${tan}^{-1}\left(x\right)\in [-\frac{\pi }{2},\frac{\pi }{2}]$. Then the minimum value of the function f : R → R defined by

$f\left(x\right)={\int }_0^{x{tan}^{-1}x}\frac{e^{(t-cost)}}{1+t^{2023}}dt$

is:

The correct answer is 0.

Step 1: Understand the given function

We are given the function f : ℝ → ℝ defined by:

$f\left(x\right)={\int }_0^{x{tan}^{-1}x}\frac{e^{(t-cost)}}{1+t^{2023}}dt$

Step 2: Analyze the upper limit of integration

Let $g\left(x\right)=x{tan}^{-1}x$.

For any real number x ∈ ℝ:

- If $x>0$, then ${tan}^{-1}x>0$, so $g\left(x\right)>0$.

- If $x<0$, then ${tan}^{-1}x<0$, so $g\left(x\right)=(-|x\left|\right)(-|{tan}^{-1}x\left|\right)>0$.

- If $x=0$, then $g\left(0\right)=0·0=0$.

Thus, $g\left(x\right)\ge 0$ for all $x\in ℝ$, and its minimum value is $0$, occurring uniquely at $x=0$.

Step 3: Analyze the integrand

The integrand is given by:

$h\left(t\right)=\frac{e^{(t-cost)}}{1+t^{2023}}$

For any $t\ge 0$, the exponential term $e^{(t-cost)}>0$ and the denominator $1+t^{2023}\ge 1>0$. Therefore, $h\left(t\right)>0$ for all $t\ge 0$.

Step 4: Determine the minimum value of f(x)

Since the integrand $h\left(t\right)$ is strictly positive for $t\ge 0$, integrating $h\left(t\right)$ from $0$ to a non-negative upper limit $g\left(x\right)$ gives:

$f\left(x\right)={\int }_0^{g\left(x\right)}h\left(t\right)dt\ge 0$

The integral evaluates to $0$ if and only if the upper limit of integration equals the lower limit of integration, i.e., when $g\left(x\right)=0$.

At $x=0$:

$f\left(0\right)={\int }_0^{0}h\left(t\right)dt=0$

For any $x\ne 0$, $g\left(x\right)>0$, which implies $f\left(x\right)>0$.

Thus, the minimum value of the function $f\left(x\right)$ is 0.