Trace $g(x)$

Calculus Level 3

$\large \displaystyle g(x)= \int_0^x f(t) \, dt$

Let $g$ and $f$ be two real -valued functions satisfying the condition given above. The function $f$ has a real root on $[-1,1]$ and is also strictly increasing on this interval.

What can be said about $g$ in the given interval?

Function

g

has a local minima Function

g

has a local maxima Nothing can be said

1 solution

Kushal Bose
Aug 5, 2016

Using Newton-Leibnitz Theorem

$g'(x)=f(x)$

Let us assume a real $c$ in interval $[-1,1]$ such that $f(c)=0$

So, $g'(c)=0$

Now $g(x)$ has a extremum in this interval.

Again $g''(x)=f'(x)$

As stated that $f(x)$ is stricly increasing in this interval so $f'(x)>0$

Then $g''(c)=f'(c)>0$

So. $g(x)$ has a local minima in the interval $[-1,1]$

Can you apply Newton-Leibniz Theorem if $f$ is not continuous (continuity is not on your assumptions) ?

Gerónimo Rojas Barragán - 4 years, 10 months ago

The theorem holds almost everywhere. But, yes we need continuity at least at the root of $f$ .

A Former Brilliant Member - 4 years, 8 months ago

Take $f(x)=x+1$. Then $g(x) =[\dfrac{t^2}{2}+t]|_{t=0}^{t=x}=\dfrac{x^2}{2}+x=x(\dfrac{x}{2}+1)$ is increasing in the interval [-1, 1]. We need the root of f in $(-1, 1)$.

Julio Alejo Ruiz - 1 year, 2 months ago

How can I use LaTeX?

Julio Alejo Ruiz - 1 year, 2 months ago

Trace g ( x ) g(x) g ( x )

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Trace $g(x)$