Animated Solution for Mathematics - Differentiation: The sum of the absolute minimum and the absolute maximum values of the function $f(x)=|3x-x^2+2|-x$ in the interval $[-1,2]$ is :

We are given the function $f(x)=|3x-x^2+2|-x$ defined on the interval $[-1,2]$. The presence of the modulus sign indicates that the function is non-differentiable at points where the inner expression changes sign.

To find these breakpoints, we solve:

Using the quadratic formula, the roots are:

We must restrict our analysis to the interval $[-1,2]$. Since $\sqrt{17}\approx 4.12$, the root $x=\frac{3+4.12}{2}\approx 3.56$ lies outside our domain and is discarded.

The relevant root is $\alpha =\frac{3-\sqrt{17}}{2}\approx -0.56$, which lies within $[-1,2]$. This splits our function into two distinct pieces:

For $x\in [-1,\alpha )$, the expression inside the modulus is negative:

For $x\in [\alpha ,2]$, the expression inside the modulus is positive:

In the first interval, the derivative is $f_1^{\prime }(x)=2x-4$. Since $x<\alpha <2$, the derivative is always negative, meaning $f_1(x)$ is strictly decreasing. The maximum occurs at the left endpoint:

In the second interval, the derivative is $f_2^{\prime }(x)=-2x+2$. Setting $f_2^{\prime }(x)=0$ yields $x=1$, which is within the interval. The value at this local maximum is:

We must also check the right endpoint $x=2$:

Finally, we evaluate at the breakpoint $\alpha$. Using the identity ${\alpha }^2-3\alpha -2=0$, we find:

Comparing our candidates $\{3,2,\frac{\sqrt{17}-3}{2}\}$, the absolute maximum is $3$ and the absolute minimum is $\frac{\sqrt{17}-3}{2}$.

The sum of the absolute maximum and absolute minimum is: