Animated Solution for Mathematics - Differentiation: Let $f(x)=\{\begin{array}{ll}{e}^x,& 0\le x\le 1\\ 2-e^{x-1},& 1<x\le 2\\ x-e,& 2<x\le 3\end{array}$ and $g(x)={\int }_0^{x}f(t)dt,x\in [1,3]$ then $g(x)$ has

Imagine you are standing on a vast, flat plain, and you are tasked with tracking the total amount of water that has flowed into a reservoir. The rate of flow is given by a function $f(t)$, and the total volume of water in the reservoir at any time $x$ is given by the integral $g(x)={\int }_0^{x}f(t)dt$.

This is exactly the scenario we are facing with our function $g(x)$. We are not just doing algebra; we are tracking the accumulation of change. To understand the peaks and valleys of this accumulation, we need to look at the rate of change itself.

The Fundamental Theorem of Calculus

Many students feel the urge to immediately integrate the piecewise function $f(x)$ to find $g(x)$. While that is mathematically possible, it is like trying to measure the height of a mountain by counting every single grain of sand on it.

We have a much more elegant tool: The Fundamental Theorem of Calculus. It tells us that the derivative of an integral function is simply the integrand itself. So, we have the beautiful, simple relationship:

This is our detective's magnifying glass. To find the local maxima and minima of $g(x)$, we do not need to know the exact volume of water; we only need to know when the flow rate $f(x)$ is zero. These are our critical points.

Our function $f(x)$ is defined in three distinct stages. For $x\in [0,1]$, $f(x)=e^x$. Since $e^x$ is always positive, the flow rate is always positive, meaning the volume $g(x)$ is always increasing. No extrema exist here.

Now, let us look at the second stage, $x\in (1,2]$, where $f(x)=2-e^{x-1}$. We set this to zero:

Taking the natural logarithm of both sides, we get $x-1=\ln 2$, or:

This is our first critical point. In the third stage, $x\in (2,3]$, we have $f(x)=x-e$. Setting this to zero gives us:

Since $e\approx 2.718$, this point also falls within our interval.

The Heartbeat of the Function

Now, we must determine the nature of these points. At $x=1+\ln 2$, we check the sign of $g^{\prime }(x)=f(x)$.

For $x<1+\ln 2$, $e^{x-1}<2$, so $f(x)>0$. For $x>1+\ln 2$, $e^{x-1}>2$, so $f(x)<0$. The derivative changes from positive to negative, which is the classic signature of a local maximum.

At $x=e$, we look at $f(x)=x-e$. For $x<e$, $x-e<0$. For $x>e$, $x-e>0$.

The derivative changes from negative to positive. This is the signature of a local minimum. The reservoir was draining, and then it started filling up again.

We have successfully mapped the behavior of $g(x)$ without ever needing to perform a complex integration. We found a local maximum at $x=1+\ln 2$ and a local minimum at $x=e$.

This problem is a perfect reminder that in calculus, understanding the relationship between a function and its derivative is far more powerful than brute-force calculation. You have navigated the piecewise landscape, identified the critical points, and used the First Derivative Test to reveal the hidden geometry of the function.