Question

Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^3+ax^2+b \log_e |x|+1, x\neq 0$.

Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $[-2, -\frac{1}{2}]$. Then $|M+m|$ is equal to (Take $\log_e 2=0.7$):

Answer 1

21.1

Answer 2

1. Differentiate the function:

Given

$$f(x)=x^3+ax^2+b\ln|x|+1,\quad x\neq0,$$

its derivative is

$$f'(x)=3x^2+2ax+\frac{b}{x}.$$

2. Find parameters using the critical points:

For $x=-1$:

$$f'(-1)=3(1)+2a(-1)+\frac{b}{-1}=3-2a-b=0 \quad\Rightarrow\quad b=3-2a.$$

For $x=2$:

$$f'(2)=3(4)+2a(2)+\frac{b}{2}=12+4a+\frac{b}{2}=0.$$

Substitute $b=3-2a$:

$$12+4a+\frac{3-2a}{2}=0 \quad\Rightarrow\quad 24+8a+3-2a=0\quad\Rightarrow\quad 27+6a=0,$$

Thus,

$$a=-\frac{27}{6}=-\frac{9}{2},$$

and

$$b=3-2\left(-\frac{9}{2}\right)=3+9=12.$$

3. Substitute into $f(x)$.

Now,

$$f(x)=x^3-\frac{9}{2}x^2+12\ln|x|+1.$$

4. Evaluate $f(x)$ at the endpoints of $[-2,-\frac{1}{2}]$ and at the internal critical point $x=-1$.

• At $x=-2$:

  $$f(-2)=(-2)^3-\frac{9}{2}(-2)^2+12\ln2+1.$$

  Calculation:

  $$(-2)^3=-8,\quad (-2)^2=4,\quad \frac{9}{2}\times 4=18,$$ $$f(-2)=-8-18+12\ln2+1=-25+12\ln2.$$

  Given $\ln2=0.7$,

  $$f(-2)=-25+8.4=-16.6.$$

• At $x=-1$ (critical point inside the interval):

  $$f(-1)=(-1)^3-\frac{9}{2}(-1)^2+12\ln1+1.$$

  Since $\ln1=0$:

  $$f(-1)=-1-\frac{9}{2}+0+1=-\frac{9}{2}=-4.5.$$

• At $x=-\frac{1}{2}$:

  $$f\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^3-\frac{9}{2}\left(-\frac{1}{2}\right)^2+12\ln\frac{1}{2}+1.$$

  Calculation:

  $$\left(-\frac{1}{2}\right)^3=-\frac{1}{8},\quad \left(-\frac{1}{2}\right)^2=\frac{1}{4},\quad \frac{9}{2}\times\frac{1}{4}=\frac{9}{8},$$

  and since $\ln\frac{1}{2}=-\ln2=-0.7$:

  $$f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12(-0.7)+1=-\frac{10}{8}-8.4+1.$$ $$-\frac{10}{8}=-1.25,\quad \text{so } f\left(-\frac{1}{2}\right)=-1.25-8.4+1=-8.65.$$

5. Determine the absolute minimum and maximum on the interval:

From the computed values:

$$f(-2)=-16.6,\quad f(-1)=-4.5,\quad f\left(-\frac{1}{2}\right)=-8.65.$$

Thus, the absolute maximum is $M=f(-1)=-4.5$ and the absolute minimum is $m=f(-2)=-16.6$.

6. Compute $|M+m|$.

$$M+m=-4.5+(-16.6)=-21.1\quad\Rightarrow\quad |M+m|=21.1.$$