Question

1. (15 points) find the absolute extrema of $f(t) = t\sqrt{4 - t^2}$ on $-1, 2$.

Explanation:

Step1: Find the derivative of \( f(t) \)

We use the product rule \( (uv)^\prime = u^\prime v + uv^\prime \), where \( u = t \) and \( v=\sqrt{4 - t^2}=(4 - t^2)^{\frac{1}{2}} \).
First, \( u^\prime = 1 \).
For \( v^\prime \), use the chain rule: \( v^\prime=\frac{1}{2}(4 - t^2)^{-\frac{1}{2}}\times(-2t)=\frac{-t}{\sqrt{4 - t^2}} \).
Then by product rule:
\( f^\prime(t)=1\times\sqrt{4 - t^2}+t\times\frac{-t}{\sqrt{4 - t^2}}=\frac{4 - t^2 - t^2}{\sqrt{4 - t^2}}=\frac{4 - 2t^2}{\sqrt{4 - t^2}} \).

Step2: Find critical points

Set \( f^\prime(t) = 0 \) or \( f^\prime(t) \) undefined.

• For \( f^\prime(t)=0 \): \( \frac{4 - 2t^2}{\sqrt{4 - t^2}} = 0 \), numerator \( 4 - 2t^2 = 0\Rightarrow t^2 = 2\Rightarrow t=\pm\sqrt{2} \).
• For \( f^\prime(t) \) undefined: denominator \( \sqrt{4 - t^2}=0\Rightarrow 4 - t^2 = 0\Rightarrow t=\pm2 \).

Now, consider the interval \([-1,2]\). So critical points in the interval are \( t = \sqrt{2} \) (since \( -\sqrt{2}\approx - 1.414<-1 \), \( t = 2 \) is an endpoint).

Step3: Evaluate \( f(t) \) at endpoints and critical points

• At \( t=-1 \): \( f(-1)=-1\times\sqrt{4 - (-1)^2}=-1\times\sqrt{3}=-\sqrt{3}\approx - 1.732 \).
• At \( t = \sqrt{2} \): \( f(\sqrt{2})=\sqrt{2}\times\sqrt{4 - (\sqrt{2})^2}=\sqrt{2}\times\sqrt{2}=2 \).
• At \( t = 2 \): \( f(2)=2\times\sqrt{4 - 2^2}=2\times0 = 0 \).

Step4: Compare the values

We have \( f(-1)=-\sqrt{3}\approx - 1.732 \), \( f(\sqrt{2}) = 2 \), \( f(2)=0 \).
So the minimum value is \( -\sqrt{3} \) (at \( t=-1 \)) and the maximum value is \( 2 \) (at \( t=\sqrt{2} \)).

Answer:

Absolute minimum: \( -\sqrt{3} \) (at \( t = - 1 \)); Absolute maximum: \( 2 \) (at \( t=\sqrt{2} \))