# Question 19863

Oct 26, 2017

Here you go.

#### Explanation:

The critical point(s) of any function $g$ occur at $x = c$ when $g ' \left(c\right) = 0$. Let's consider a critical point $x = a$ for the function $f$, that is $f ' \left(c\right) = 0$. The coordinates of this critical point is given by $\left(c , f ' \left(c\right)\right)$.

Consider the function $g \left(x\right) = a f \left(x\right)$. Its critical point occurs when $g ' \left(x\right) = 0$, that is $a f ' \left(x\right) = 0$. Either $a = 0$, which is false, or $f ' \left(x\right) = 0$, which is true at least for some point $x = c$ i.e. $f ' \left(c\right) = 0$. The coordinates of this critical point therefore is given by $\left(c , g \left(c\right)\right)$, or in terms of $f$, $\left(c , a f \left(x\right)\right)$.

Consider the function $g \left(x\right) = a + f \left(x\right)$. Its critical point occurs when $g ' \left(x\right) = 0$, that is $0 + f ' \left(x\right) = 0$, which is true at least for some point $x = c$ i.e. $f ' \left(c\right) = 0$. The coordinates of this critical point therefore is given by $\left(c , g \left(c\right)\right)$, or in terms of $f$, $\left(c , a + f \left(x\right)\right)$.

Consider the function $g \left(x\right) = f \left(a x\right)$. Its critical point occurs when $g ' \left(x\right) = 0$, that is $a f ' \left(a x\right) = 0$. Either $a = 0$, which is false, or $f ' \left(a x\right) = 0$. We know that $f ' \left(c\right) = 0$, so $a x = c$ for some $c$, and $x = \setminus \frac{c}{a}$. That is, $g ' \left(\setminus \frac{c}{a}\right) = 0$ The coordinates of this critical point therefore is given by ((\frac{c}{a},g((\frac{c}{a})), or in terms of $f$, ((\frac{c}{a}, f(c))#.

Consider the function $g \left(x\right) = f \left(a + x\right)$. Its critical point occurs when $g ' \left(x\right) = 0$, that is $f ' \left(a + x\right) = 0$, which is true at least for some point $a + x = c$ i.e. $f ' \left(c\right) = 0$. Thus, $x = c - a$ and $g ' \left(c - a\right) = 0$. The coordinates of this critical point therefore is given by $\left(c - a , g \left(c - a\right)\right)$, or in terms of $f$, $\left(c - a , f \left(c\right)\right)$.