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why this function $x^{4} + y^{4} - 4 x y + 1$ graph differently between mathematica and my textbook? did I make some error with my mathematica syntax?

enter image description here

enter image description here

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    What scale are we seeing for the textbook? – 2017-02-01
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    Take smaller range of $x,y$, e.g. $-2$ to $2$. – 2017-02-01
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    Impossible to answer if no axis values are displayed in both figures – 2017-02-01
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    Look at the scales on the Mathematic plot. The $z$-direction ticks are multiples of $100000$. Compare with the other direction. You will find that the $z$-direction is compressed by a factor of $10000$. Any smaller local variation is drowned. And, no this is not Mathematica's fault, you asked it to plot this function in this area, and it is doing the best to cope. – 2017-02-01

2 Answers 2

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If we let $f(x,y)=x^4-4 x y+y^4+1$ we can use the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ to calculate that those humps that you see in the picture from the book must be at the points $(1,1,-1)$ and $(-1,-1,-1)$. So now we know that we should take something like $-2$ to $2$ for our $x$ and $y$ range. By evaluating $f(0,0)=1$ we see that we should not take our $z$ range to have a maximum value that is too big.

Try something like 

Plot3D[x^4 + y^4 - 4 x*y + 1, {x, -2, 2}, {y, -2, 2}, PlotRange ->
{All, All, {-1, 2}}]

which results in

enter image description here

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If $f(x,y)=x^4+y^4-4xy+1$, then $$ D_1f(x,y)=4x^3-4y\\ D_2f(x,y)=4y^3-4x $$ which has critical points at $(0,0)$, $(-1,-1)$ and $(1,1)$.

The Hessian is $H(x,y)=16(9x^2y^2-1)$; since $H(0,0)<0$ and $H(-1,-1)=H(1,2)>0$, we have that $(0,0)$ is a local maximum, while $(-1,-1)$ and $(1,1)$ are local minimum.

Now $f(0,0)=0$ and $f(-1,-1)=f(1,1)=-1$. With your $z$-axis scaled to fit the interval $(0,300\,000)$, the difference between $0$ and $-1$ cannot be appreciated.