# Global min/max?

the function f(t)= t/5+t^(2)

I got sqrt 5/10 for the max is that correct?

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• TomV
Lv 7
4 weeks ago

f(t) = t/5 + t²

f'(t) = 1/5 + 2t

A local min/max occurs when f'(t) = 0 → t = -1/10

f(-1/10) = -1/50 + 1/100 = -1/100

A global extrema is the extreme value over the domain of the function. Since an explicit domain is not specified for the function, the implicit domain is assumed. The implicit domain of f(t) = t/5 + t² is (-∞, ∞)

lim(x→-∞) f(t) = ∞

lim(x→∞) f(t) = ∞

The global max of the function would be lim(x→±∞) f(x) = ∞

The global min of the function would be f(-1/10) = -1/100

sqrt 5/10 is not an extreme value of the function you presented.

• rotchm
Lv 7
4 weeks ago

There are many ways to tackle this. Here is one way:

Recalling your highschool math, notice that your function is a second deg poly, a parabola. You noticed that?  Its also an upwards parabola (why?). It thus has a min, but no max (you have not specified a range, thus the range is all t's.).

Just recall the formula for finding the min (the vertex). Or, you know the vertex of a parabola is midpoint of the roots. Here, the roots can be found this way: Rewrite your fct as t(1/5+t). The roots are thus 0 & -1/5. Thus the x_vertex is midway, -1/10.

Thus, the y_vertex is f(-1/10) = [left for you to compute]. What do you get?

Done!

No calculus involved in this procedure. Only highschool math!

• Philip
Lv 6
4 weeks ago

f(t) = (1/5)t + t^2, f'(t) = (1/5) + 2t, f''(t) = 2, > 0.;

Extremums occur where f'(t) = 0. Here, f'(t) = 0 ---> t = (-1/10). Since f''(t) > 0 for all t,

any extremum will be a minimum. At t = (-1/10), f(t) = (1/5)(-1/10)+(-1/10)^2 = -0.01.

f(t) is a concave upwards parabola because the t^2 coefficient is positive. Recall that f''(t) indicates the rate of change for f'(t) as t increases. A positive rate of change

occurs when f'(t) becomes less negative or more positive as t increases. It indicates

that the curve is concave upwards and only a minimum can occur. A negative rate of change occurs when f'(t) becomes more negative or less positive as t increases. It

indicates that the curve is concave downwards and only a maximum can occur. This

is why the test using f''(t) is essential in determining whether a min or max occurs at

a critical point.

• 4 weeks ago

A local min or max of a function will occur when the derivative at that point equals zero. So let's take a derivative, find what t makes it zero, and then plug it back in.

f'(t) = 2t + 1/5 = 0

t = -1/10

f(-1/10) = -1/50 + 1/100 = -1/100

Now, we can take the second derivative to check if it's a min or a max.

f''(t) = 2 > 0 for all t, so it must be a minimum. (You can check this via a graph)

In this case the tail behaviour is also up, so we can say that the function has a global minimum of  -1/100 at t = -1/10.