# Math Help - Global max/min.

1. ## Global max/min.

im having trouble with these two, finding the global max/min, i just dont understand how to do it...

g(t) = t/(t^2+5)

Determine the global extrema of g(t) on the interval 1 ≤ t ≤ 8.

g(t) has a global minimum at t =

g(t) has a global maximum at t =

f(theta) = sin^2(theta)-cos(theta)

Determine the global extrema of f(θ) on the interval 0 ≤ θ ≤ pi.

f(θ) has a global maximum at θ =

f(θ) has a global minimum at θ =

2. Hello,

You can derivate the functions and see where the derivate function has zeros and see which one is a maximum or a minimum ~

3. Originally Posted by mathlete
im having trouble with these two, finding the global max/min, i just dont understand how to do it...

g(t) = t/(t^2+5)

Determine the global extrema of g(t) on the interval 1 ≤ t ≤ 8.

...
From the equation of

$g(t)=\frac t{t^2+5}$

you can see that

g(t) > 0 if t > 0
g(t) < 0 if t < 0 (keep in mind that $t^2+5 >=5~,~t \in \mathbb{R}$

$\lim_{|t| \mapsto \infty} = 0$ and therefore the graph of g approaches the t-axis from above for t > 0 and it approaches the t-axis from below for t < 0.

Calculate g'(t):

$g'(t) = \frac{(t^2+5)-t \cdot 2t}{(t^2+5)^2}$

g'(t) = 0 if the numerator equals zero:

$(t^2+5)-t \cdot 2t = 0$ I've got $|t|=\sqrt{5}$

Use my considerations to determine maximum or minimum points of the graph.

(Remark: The axes are not scaled equally!!!)

4. yeah i know how to find local max/min, i already got those, but the global is what gives me trouble, i just dont know how to find it...

5. Global maximum means the maximum in the entire interval. The global maximum will be among the local maximums of the interval.

Normally, you shouldn't have more than 2 extrema for these functions.

Study the variations of the functions to determine which one is a max or min

6. A differentiable function will achieve a critical point at those values where its derivative vanishes.

$g(t) = \frac{t}{t^2+5}$

$g'(t) = \frac{(t^2+5)(1)-(t)(2t)}{(t^2+5)^2} = 0$

$g'(t) = \frac{5-t^2}{(t^2+5)^2} = 0$ when $t= \pm \sqrt{5}$

There may be some cleverness in determining which point is max and which is a min or you can take the second derivative and determine the concavity of the function at that point. Now these are local max and mins (or critical points), but the global minimum may be different since the function may approach a smaller or larger value for large values of $|t|$. In the case of the above function the limiting behavior is $g(t) \rightarrow 0$ as $t \rightarrow \pm \infty$ and the local extrema found above are in fact global extrema.

But take for example $f(x) = \frac{e^x}{x^2+\frac{1}{2}}$.

$f'(x) = \frac{(x^2+\frac{1}{2})e^x - e^x(2x)}{(x^2+\frac{1}{2})^2}$

which vaishes when $x = \frac{2 \pm \sqrt{2}}{2}.$These points are the local max and min of the function but are not the global max or min. The function does not achieve a global minimum as it asymptotically approaches 0, so the infimum of the function is 0 but the function never actually acheives that value. And the function is unbounded as $x \rightarrow \infty$ meaning it has no max.

To prove the local min is the global min you can argue that the function is bounded by the suspected global minimum. In your case show $g(t) \geq g(-\sqrt{5}) = \frac{-\sqrt{5}}{10}, \forall t$.