Find the absolute maxima and absolute minima of the function f(x) = x^3 3x^2 + 5 on [ 2, 2] . Also find the point of inflection. [4+1]

Solution

Step 1: Find the first derivative and critical points

To find the absolute maxima and minima of the function $f(x)=x^3-3x^2+5$ on the interval $[-2,2]$, we first find its critical points by differentiating $f(x)$ and setting the derivative equal to zero.

$$f^{\prime }(x)=\frac{d}{dx}(x^3-3x^2+5)=3x^2-6x$$

Set $f^{\prime }(x)=0$:

$$\begin{gathered} 3x^2-6x=0 \\ 3x(x-2)=0 \\ \Rightarrow x=0\text{ or }x=2 \end{gathered}$$

So, the critical points are at $x=0$ and $x=2$.

Step 2: Evaluate $f(x)$ at critical points and endpoints

We evaluate $f(x)$ at the critical points and the endpoints of the interval $[-2,2]$:

$$\begin{array}{cc}x& f(x)\\ -2& f(-2)=(-2{)}^3-3(-2{)}^2+5=-8-12+5=-15\\ 0& f(0)=0^3-3(0{)}^2+5=5\\ 2& f(2)=2^3-3(2{)}^2+5=8-12+5=1\end{array}$$

Step 3: Compare values to find absolute extrema

From the table above, the values of $f(x)$ at the critical points and endpoints are:

• $f(-2)=-15$
• $f(0)=5$
• $f(2)=1$

Additionally, we check the value at $x=1$ (since it is a point of interest in the interval):

$$f(1)=1^3-3(1{)}^2+5=1-3+5=3$$

Now, comparing all these values:

• The absolute maximum value of $f(x)$ on $[-2,2]$ is 5 at $x=0$.
• The absolute minimum value of $f(x)$ on $[-2,2]$ is -15 at $x=-2$.

Step 4: Find the point of inflection

To find the point of inflection, we compute the second derivative and set it to zero:

$$f^{\prime \prime }(x)=\frac{d}{dx}(3x^2-6x)=6x-6$$

Set $f^{\prime \prime }(x)=0$:

$$\begin{gathered} 6x-6=0 \\ 6x=6 \\ x=1 \end{gathered}$$

Now, verify that this is indeed a point of inflection by checking the concavity on either side of $x=1$:

• For $x<1$ (e.g., $x=0$): $f^{\prime \prime }(0)=-6<0$ (concave down).
• For $x>1$ (e.g., $x=2$): $f^{\prime \prime }(2)=6>0$ (concave up).

Since the concavity changes at $x=1$, the point of inflection is at $x=1$. The corresponding $y$-value is:

$$f(1)=3$$

Thus, the point of inflection is at (1, 3).