If $f(x,y)=x^2+y$, what is the image of $K=\{(x,y):x^2+y^2\leq 1\}$? Please disregard the first eight lines of the solution below (which I have provided for completeness; the referenced theorems simply state that continuous functions between metric spaces preserve compactness and connectedness).
My question concerns only lines 5 and 3 from the bottom. In each of these cases, while $t$ certainly equals $f(x,y)$ for some $(x,y) \in \mathbb R^2$, the $x^2+y^2\in[1,2]$ so that $(x,y) \notin K$-- i.e., contrary to the author's assertion, we have not shown that $t\in f(K)$, right?

Daniel has pointed out in the comments that $f(K)$ is actually $[-1,\frac 54]$.
How to show that $t\in (1,\frac 54]\implies t\in f(K)$? And that $t\in f(K)\implies t\in [-1,\frac 54]$?


P.S. This exercise is from Marsden's Elementary Classical Analysis.
 A: One way to see that $f(K)=[-1,5/4]$ is to first notice that the maximum and minimum values of $f$ cannot occur inside the circle, since $f_{x}=2x$ and $f_{y}=1$ are not both zero anywhere inside the circle.  (More simply, a maximum cannot occur at a point $(x,y)$ inside the circle, since $f(x,y+h)>f(x,y)$ if $0<h<\sqrt{1-x^2}-y$; and similarly for a minimum.)
On the boundary, letting $x=\cos\theta$ and $y=\sin\theta$ gives the function
$g(\theta)=\cos^{2}\theta+\sin\theta$ for $0\le \theta\le 2\pi$.
Then $g^{\prime}(\theta)=2(\cos\theta)(-\sin\theta)+\cos\theta=\cos\theta(1-2\sin\theta)=0$ if $\cos\theta=0$ or $\sin\theta=\frac{1}{2}$.
If $\cos\theta=0$, then $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$; and if $\sin\theta=\frac{1}{2}$, $\theta=\frac{\pi}{6}$ or $\theta=\frac{5\pi}{6}$.
Since $g(\frac{\pi}{2})=1, \;\;g(\frac{3\pi}{2})=-1,\;\; g(\frac{\pi}{6})=\frac{5}{4}\;\;$, and $\;\;g(\frac{5\pi}{6})=\frac{5}{4}$,
$g$ has a maximum of $\frac{5}{4}$ and a minimum of -1 on $[0,2\pi]$.
By the Intermediate Value Theorem, $g([0,2\pi])=[-1,\frac{5}{4}]$; and
therefore
$f(K)=[-1,\frac{5}{4}]$. 
A: Here is a calculational solution which does not use goniometry.  As in the question, all variables range over $\mathbb R$.
In essence, we're asked to determine the set which, for each $\;x,y\;$ such that $\;x^2+y^2\leq 1\;$, contains $\;x^2+y\;$.  Or in Dijkstra's notation (see EWD1300 under quantifications) we're asked to simplify the set $$\;\langle x,y : x^2+y^2 \leq 1 : x^2+y \rangle\;$$
Therefore we investigate which elements this set has, and try to simplify along the way: for any $\;z\;$,
\begin{align}
& z \in \langle x,y : x^2+y^2 \leq 1 : x^2+y \rangle \\
\equiv & \;\;\;\;\;\text{"definition of set comprehension"} \\
& \langle \exists x,y : x^2+y^2 \leq 1 : x^2+y = z \rangle \\
\equiv & \;\;\;\;\;\text{"rearrange -- to prepare for one-point rule"} \\
& \langle \exists x,y : y = z-x^2 : x^2+y^2 \leq 1 \rangle \\
\equiv & \;\;\;\;\;\text{"one-point rule"} \\
& \langle \exists x :: x^2+(z-x^2)^2 \leq 1 \rangle \\
\equiv & \;\;\;\;\;\text{"abbreviate common $\;x^2\;$ by $\;w\;$"} \\
& \langle \exists w : \langle \exists x :: w = x ^ 2 \rangle : w+(z-w)^2 \leq 1 \rangle \\
\equiv & \;\;\;\;\;\text{"simplify; expand square and rewrite as polynomial in $\;w\;$"} \\
& \langle \exists w : w \geq 0 : w^2+(1-2z)w+(z^2-1) \leq 0 \rangle \\
\end{align}
We have arrived at the quadratic form for a parabola which opens up, and we're asked for the values $\;\geq 0\;$ for which it is $\;\leq 0\;$.  A quick sketch shows that this occurs whenever the positive root exists and is $\;\geq 0\;$, and therefore we continue
\begin{align}
\equiv & \;\;\;\;\;\text{"discriminant $\;\geq 0\;$ and positive root $\;\geq 0\;$"} \\
& (1-2z)^2-4(z^2-1) \geq 0 \;\land\; \frac{2z-1 + \sqrt {(1-2z)^2-4(z^2-1)}}{2} \geq 0 \\
\equiv & \;\;\;\;\;\text{"simplify left and right part"} \\
& 5-4z \geq 0 \;\land\; \sqrt {5-4z} \geq 1-2z \\
\equiv & \;\;\;\;\;\text{"left hand part: simplify; right hand part: square both sides"} \\
& z \leq 5/4 \;\land\; (1-2z < 0 \:\lor\: (5-4z)^2 \geq (1-2z)^2) \\
\equiv & \;\;\;\;\;\text{"right hand part: simplify"} \\
& z \leq 5/4 \;\land\; (z > 1/2 \:\lor\: -1 \leq z \leq 1) \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& -1 \leq z \leq 5/4 \\
\equiv & \;\;\;\;\;\text{"rewrite as a set -- since this was our goal all along"} \\
& z \in [-1, \frac 5 4] \\
\end{align}
Therefore the image is $\;[-1, \frac 5 4]\;$.
