Why is the maximum acceleration must occur at the extreme ends of a simple harmonic motion? Description of the diagram in my textbook's exact words:

"The following acceleration-time graph shows the motion of a particle
with initial displacement at the centre of motion where acceleration
is zero."

It then adds:

"The maximum acceleration occurs at the extreme ends of the motion"


What does it mean by "extreme points", is that referring to end points? So i suppose, $t=0$ and $t=T$?
 A: A harmonic motion $t\mapsto x(t)$ is characterized by the ODE
$$\ddot x(t)\equiv-\omega^2 x(t)\ .\tag{1}$$
This equation results from the idea that the repulsing force is proportional to the deviation $|x(t)|$ from the equivalence point, and Newton's law connecting the force with the resulting acceleration.
Inspecting $(1)$ one sees immediately that the acceleration $\ddot x(t)$ is minimal (as negative as possible) at the moments $t$ where the deviation $x(t)$ is maximal (i.e. at the upper extreme), and is maximal possible, when the deviation is at its minimum (i.e. at the lower extreme).
A: Maximum displacement (amplitude ) and acceleration  occur at the ends of displacement i.e., maximum/minimum amplitude where velocity is zero,i.e., at $t= T/4, 3 T/4$
We can also plot simple harmonic motion dynamics between velocity and acceleration in order to see in the ellipse ends of major/minor axes where maxima/minima occur.
$$ \big(\dfrac{\dot x}{v_{max}} \big)^2+\big(\dfrac{\ddot x} {v_{max}^2/A} \big)^2= 1 $$
where
$$v_{max,min} = \pm \omega A;\; Accln_{max,min} = \pm \omega^2 A  $$

A: This is acceleration vs time graph. So on y-axis is value of acceleration. For maximum/minimum values one has to look at the peak values of graph. But you're looking at the slope of the acceleration-time graph!
A: The physical answer is that if the oscillator must be pulled most strongly exactly when it is at the farthest from the equilibrium. If not, where else would you think the pull would be greatest?
In a purely mathematically perspective, an oscillator's equation has the form:
$$ \ddot{x} = -\kappa x$$
this suggests that, the acceleration is directly a function of where the particle is, so the first argument I told could easily be made into the argument using equations by just replacing a few words.
