Studying the monotonicity of $f(x) = 1- (x-5)^2$ without using differentiation and continuity.

Question

How can I study the monotonicity of $f(x) = 1- (x-5)^2$ without using differentiation and continuity? and exactly by the method like dividing the domain into intervals (how I will divide the $\mathbb{R}$ into intervals? depending on what? )and studying the behavior of $f$ in each interval, could anyone help me please?

Answer 1

$x^2$ is an upward parabola with vertex at $(0,0)$.

$-x^2$ is a downward parabola with vertex at $(0,0)$.

$1-x^2$ is a downward parabola with vertex at $(0,1)$.

$1-(x-5)^2$ is a downward parabola with vertex at $(5,1)$.

Therefore, $f(x)=1-(x-5)^2$ increases on $(-\infty,5]$ and decreases on $[5,\infty)$.

Answer 2

Let's have $y\ge x$ or $y=x+u$ with $u\ge 0$ then

$f(y)-f(x)=(x-5)^2-(y-5)^2=(x-5-y+5)(x-5+y-5)=(x-y)(x+y-10)$

• $f(y)-f(x)=-u(2(x-5)+u)$

If $y\ge x\ge 5$ then $(x-5)\ge 0,u\ge 0$ so $f(y)-f(x)\le 0$ and $f\searrow$

• $f(y)-f(x)=u(u+2(5-y))$

If $x\le y\le 5$ then $(5-y)\ge 0,u\ge 0$ so $f(y)-f(x)\ge 0$ and $f\nearrow$

Answer 3

Part 0.

Begin by proving facts like: "If $f$ is order-preserving on $A$ and $g$ is order-preserving on $f(A)$, then $g \circ f$ is order-preserving on $A$."

There's three other such facts that can be obtained by replacing the word 'preserving' with the word 'reversing' in appropriate combinations.

Part 1.

Show that the function $\Box^2$ defined by $\Box^2(x) = x^2$ is decreasing on $(-\infty,0]$ and increasing on $[0,\infty)$.

Part 2. Express $f$ as $$(1-\Box) \circ \Box^2 \circ (\Box -5)$$ with the obvious meanings.

Part 3. Find appropriate sets $A \subseteq \mathbb{R}$ on which to verify that the above composite is order-preserving or order-reversing.

Answer 4

I made a mistake when looking at your function at first glance and gave a geometric explanation for a slightly different function. But I think it might help anyway.

Consider $$ g(x)=1+(x-5)^2=(0-1)^2+(x-5)^2 $$ and note the distance between two points $(x_1, y_1), (x_2, y_2)$ in the plane is $$ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} $$ let $(x_2, y_2)=(1, 5)$ and $(x_1, y_1)=(x, 0)$

then $(x_1, y_1)$ is points in the x-axis, and $g(x)$ measures the distance (indeed, the square of distance) from it to a fixed point $(1, 5)$

as x moving from $-\infty$ to 5, the distance decrease monotonically; and then increases monotonically when x moves from 5 to $\infty$; so achieve minimum at x=5