Find X For Twice-Differentiable Function F

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into some advanced calculus, specifically for those of you tackling AP Calculus or anything similar. We've got a super interesting problem involving a twice-differentiable function, $f$, where we know $f(0)=4$. The real kicker is its derivative, $f^{\prime}(x)=\sin \left(x^2-2 x+1\right)$, and we're working within the interval $-2 \leq x \leq 2$. Our mission, should we choose to accept it, is to find all values of $x$ in this interval where the function might have local extrema. This means we're on the hunt for points where the derivative is either zero or undefined. Since $f^{\prime}(x)$ is a sine function composed with a polynomial, it's defined everywhere, so we're focusing on where $f^{\prime}(x)=0$. Let's break this down, shall we?

First off, let's simplify that derivative. Remember your algebra, guys? The expression inside the sine function, $x^2 - 2x + 1$, is a perfect square! It's $(x-1)^2$. So, our derivative becomes $f^{\prime}(x)=\sin \left((x-1)^2\right)$. Now, to find potential local extrema, we need to set this equal to zero: $\sin \left((x-1)^2\right) = 0$. The sine function equals zero when its argument is an integer multiple of $\pi$. That is, $\sin(\theta) = 0$ when $\theta = n\pi$ for any integer $n$. In our case, the argument is $(x-1)^2$, so we have $(x-1)^2 = n\pi$. We're looking for values of $x$ in the interval $[-2, 2]$. This means $(x-1)^2$ must be non-negative. Since $n\pi$ must be non-negative, $n$ must be a non-negative integer ($n=0, 1, 2, \dots$). Let's analyze this equation, $(x-1)^2 = n\pi$, for different values of $n$.

When $n=0$, we have $(x-1)^2 = 0\pi = 0$. Taking the square root of both sides gives us $x-1 = 0$, which means $x=1$. Is $x=1$ within our interval $[-2, 2]$? Absolutely! So, $x=1$ is a critical point. Now, what about $n=1$? We have $(x-1)^2 = 1\pi = \pi$. Taking the square root gives us $x-1 =

Let's continue our quest, shall we? We've established that we need to solve $(x-1)^2 = n p$ for integer values of $n$ where $n p p$ is non-negative, and $x$ falls within the interval $[-2, 2]$. We already found $x=1$ for $n=0$. Now let's explore $n=1$. The equation becomes $(x-1)^2 = p$. Taking the square root of both sides, we get $x-1 =

p$ or $x-1 = -

p$. This gives us two potential values for $x$: $x = 1 +

p$ and $x = 1 -

p$. We need to check if these values lie within our interval $[-2, 2]$. We know that $p p p 1.77$. So, $x p p p 1 + 1.77 = 2.77$ and $x p p p 1 - 1.77 = -0.77$. The value $x p p p 2.77$ is outside our interval $[-2, 2]$. However, $x p p p -0.77$ is inside our interval. So, $x = 1 -

p$ is another critical point we need to consider. What about $n=2$? The equation is $(x-1)^2 = 2 p$. Taking the square root, $x-1 =

p$ or $x-1 = -

p$. This yields $x = 1 +

p$ and $x = 1 -

p$. We know $p p p 2.50$. So, $x p p p 1 + 2.50 = 3.50$ and $x p p p 1 - 2.50 = -1.50$. The value $x p p p 3.50$ is outside our interval. However, $x p p p -1.50$ is inside our interval. Thus, $x = 1 -

p$ is another critical point. Let's check $n=3$. The equation is $(x-1)^2 = 3 p$. Taking the square root, $x-1 =

p$ or $x-1 = -

p$. This gives us $x = 1 +

p$ and $x = 1 -

p$. We know $p p p 3.06$. So, $x p p p 1 + 3.06 = 4.06$ and $x p p p 1 - 3.06 = -2.06$. The value $x p p p 4.06$ is outside our interval. The value $x p p p -2.06$ is also outside our interval $[-2, 2]$. It's very close, but outside. As $n$ increases, $(x-1)^2 = n p$ will lead to larger values of $x-1$ and consequently larger absolute values of $x$. This means for $n p 3$, the solutions for $x$ will likely fall outside our interval $[-2, 2]$. So, we only need to consider $n=0, 1, 2$.

Let's recap the critical points we've found so far within the interval $[-2, 2]$: $x=1$ (from $n=0$), $x = 1 -

p$ (from $n=1$), and $x = 1 -

p$ (from $n=2$). These are the points where $f^{\prime}(x)=0$. But hold on a sec, guys! The question asks for all values of $x$ where $f$ might have local extrema. This includes points where $f^{\prime}(x)=0$ and points where $f^{\prime}(x)$ is undefined. Since $f^{\prime}(x)=\sin \left((x-1)^2\right)$ is a continuous function everywhere (it's a composition of sine and a polynomial), there are no points where $f^{\prime}(x)$ is undefined. Therefore, our only candidates for local extrema are the points where $f^{\prime}(x)=0$. So, the values of $x$ we've identified are indeed our potential local extrema.

Let's organize these values: $x=1$, $x = 1 -

p$, and $x = 1 -

p$. It's crucial to remember that these are critical points, which are candidates for local extrema. To confirm if they are indeed local maxima or minima, we would typically use the First Derivative Test or the Second Derivative Test. The problem, however, only asks to find all values of x where it might have local extrema, which precisely means finding the critical points within the given interval. So, we've done our job! The values of $x$ in the interval $[-2, 2]$ where $f$ might have local extrema are $x=1$, $x = 1 -

p$, and $x = 1 -

p$. Pretty neat, right? Keep practicing these calculus concepts, and you'll master them in no time! Don't forget to tune in next time for more math adventures here at Plastik Magazine!

To be super thorough, let's double-check our interval constraints. We are given $-2 p x p 2$. This means $x$ must be greater than or equal to $-2$ and less than or equal to $2$. Let's re-evaluate our solutions.

For $n=0$, we got $(x-1)^2 = 0$, which gives $x=1$. Since $-2 p 1 p 2$, this value is valid.

For $n=1$, we got $(x-1)^2 = p$. This leads to $x-1 =

p$ or $x-1 = -

p$. So, $x = 1 +

p$ and $x = 1 -

p$. We know $p p p 1.772$. Thus, $x p p p 1 + 1.772 = 2.772$ and $x p p p 1 - 1.772 = -0.772$. The value $x p p p 2.772$ is outside our interval $[-2, 2]$. However, $x p p p -0.772$ is within the interval. So, $x = 1 -

p$ is a valid critical point.

For $n=2$, we got $(x-1)^2 = 2 p$. This leads to $x-1 =

p$ or $x-1 = -

p$. So, $x = 1 +

p$ and $x = 1 -

p$. We know $p p p 2.507$. Thus, $x p p p 1 + 2.507 = 3.507$ and $x p p p 1 - 2.507 = -1.507$. The value $x p p p 3.507$ is outside our interval $[-2, 2]$. However, $x p p p -1.507$ is within the interval. So, $x = 1 -

p$ is a valid critical point.

For $n=3$, we got $(x-1)^2 = 3 p$. This leads to $x-1 =

p$ or $x-1 = -

p$. So, $x = 1 +

p$ and $x = 1 -

p$. We know $p p p 3.069$. Thus, $x p p p 1 + 3.069 = 4.069$ and $x p p p 1 - 3.069 = -2.069$. Both $x p p p 4.069$ and $x p p p -2.069$ are outside our interval $[-2, 2]$.

For $n=4$, we get $(x-1)^2 = 4 p$. This leads to $x-1 =

p$ or $x-1 = -

p$. So, $x = 1 + 2

p$ and $x = 1 - 2

p$. Since $2

p p 5.013$, these values are clearly outside the interval $[-2, 2]$. As $n$ increases, the values of $

p$ increase, and thus the resulting $x$ values will move further away from the interval $[-2, 2]$. Therefore, we can stop here.

So, the values of $x$ in the interval $[-2, 2]$ where $f$ might have local extrema are precisely the critical points we found within this interval where $f^{\prime}(x) = 0$. These values are $x=1$, $x = 1 -

p$, and $x = 1 -

p$. These are the points where the tangent line to the function $f(x)$ is horizontal. They are the candidates for local maxima and minima.

Final Answer: The values of $x$ in the interval $[-2, 2]$ where $f$ might have local extrema are $x=1$, $x = 1 -

p$, and $x = 1 -

p$. Keep studying, keep exploring, and keep that mathematical curiosity alive, guys!