The function $f(x)$ is defined only on domain $[-1,2]$, and is defined on this domain by the formula
$$f(x) = 2x^2-8x+1.$$
What is the range of $f(x)$? Express your answer as an interval or as a union of intervals.
Completing the square, we obtain \begin{align*}
f(x)&=2\left(x-\frac{4}{2}\right)^2+1-2\left(\frac{4}{2}\right)^2 \\
&=2\left(x-2\right)^2+1-2(4) \\
&=2\left(x-2\right)^2-7.
\end{align*} Since $(x-2)^2\ge 0$, we have $2(x-2)^2\ge 0$. Therefore, $f(x) \ge 0 - 7 = -7$. Since $f(x)$ is a quadratic with a positive leading coefficient, we know $f(x)$ is minimized at its vertex. The vertex occurs when $x=2$, so when $x$ is in the domain $[-1,2]$, we know $f(x)$ is at most $2(0)^2 - 7 =-7$. Therefore, the range of $f(x)$ is $[-7, \infty)$.
To find the range of the function $f(x)$, we need to determine the set of possible values that $f(x)$ can take.
Step 1: Find the vertex of the parabola represented by the equation $f(x) = 2x^2 - 8x + 1$. The $x$-coordinate of the vertex is given by $x = -\frac{b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic equation.
In this case, $a = 2$ and $b = -8$, so the $x$-coordinate of the vertex is $x = -\frac{-8}{2(2)} = 2$.
Step 2: Substitute the $x$-coordinate of the vertex into the quadratic equation to find the $y$-coordinate of the vertex.
When $x = 2$, we have $f(2) = 2(2)^2 - 8(2) + 1 = 4 - 16 + 1 = -11$.
Therefore, the vertex of the parabola is $(2, -11)$.
Step 3: Notice that the coefficient of the $x^2$ term is positive ($2$), which means the parabola opens upward. This implies that the vertex is the lowest point on the graph of the function.
Step 4: The parabola will not extend beyond the domain $[-1,2]$. Therefore, the range of $f(x)$ will be from the $y$-coordinate of the vertex ($-11$) to the maximum value of $f(x)$ within the domain.
Step 5: To find the maximum value of $f(x)$ within the domain, we evaluate $f(-1)$ and $f(2)$.
For $f(-1)$, we have $f(-1) = 2(-1)^2 - 8(-1) + 1 = 2 + 8 + 1 = 11$.
Since $f(-1) = 11$ is greater than the $y$-coordinate of the vertex ($-11$), the maximum value of $f(x)$ within the domain is $f(-1)$.
Step 6: Therefore, the range of $f(x)$ is $[-11, 11]$.
To find the range of a function, we need to determine all possible values of the function output.
In this case, the function $f(x) = 2x^2 - 8x + 1$ is a quadratic function. Since the leading coefficient is positive, the parabola opens upwards. This means that the vertex of the parabola represents the lowest point of the graph, and the range of the function will extend upwards.
To find the vertex of the parabola, we can use the formula $x = -\frac{b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic function. In our case, $a = 2$ and $b = -8$. Plugging these values into the formula, we get:
$x = -\frac{-8}{2(2)} = -\frac{-8}{4} = 2$
So, the vertex of the parabola is at $(2, f(2))$. Now we can substitute this value of $x$ into the function to find the corresponding $y$-value:
$f(2) = 2(2)^2 - 8(2) + 1 = 8 - 16 + 1 = -7$
Therefore, the vertex of the parabola is $(2, -7)$.
Since the parabola opens upwards, the range of the function will extend from the $y$-value of the vertex and go infinitely upwards. So, the range of $f(x)$ is $[-7, \infty)$.