Linear approximation VS exact value for this function in $\mathbb{R}^2$

Question

Consider $f: \mathbb{R}^2\to \mathbb{R}$ to be a differentiable function.

We have $f\left(\frac{9}{10}, \frac{1}{10}\right) = 3$, $f'_x\left(\frac{9}{10}, \frac{1}{10}\right) = 1$, $f'_y\left(\frac{9}{10}, \frac{1}{10}\right) = -2$.

Using the best linear local approximation of $f$ around the point $\left(\frac{9}{10}, \frac{1}{10}\right)$, calculate an approximate value for $f(1, 0)$.

Then, assuming that $f$ is strictly concave, say if the exact value of $f(1, 0)$ is greater or lower than the approximate value.

Attempts

So for the first request, I just applied Taylor series:

$$f(x, y) = f(x_0, y_0) + (x-x_0)f'_x(x_0, y_0) + (y-y_0)f'_y(x_0, y_0) + O(||x-y||^2)$$

With $x_0, y_0$ the given point and hence I got

$$f(x, y) = x + y + 2$$

hence $$f(1, 0) \approx 3$$

Now for the second part I don't know how to move.

I thought about using the definition of concave function in two variables namely

$$f(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \geq \lambda f(x_1, y_1) + (1-\lambda)f(x_2, y_2)$$

I have no other information. I tried to chose $\lambda = 1/2$ but I got no result since I don't know the values of the function at other points.

Any help? Thank you!

Answer 1

Try to create some intuition behind concavity:

• The expression $\lambda f(x_1,x_2) + (1-\lambda)f(x_2,y_2)$ is simply the straight line connected the two values of $f$.
• The expression $f(\lambda x_1+(1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2)$ is simply $f$ applied to the straight line $\lambda (x_1,y_1) + (1-\lambda)(x_2,y_2)$.
• Intuitively, one can think of the latter as being the first, but 'projected' onto the graph of $f$.
• Now, if the first expression is greater than the second, it means we project that line upwards onto the graph. If you try to visualize, it just means that near these point, $f$ points 'downwards'.
• The best linear approximation of $f$ at a point is simply the tangent plane at that point (here $z=x+y+2)$. Given that $f$ points 'downward', the actual value will be smaller than the approximate value.

I hope it is clear that if "concave" were to be replaced by "convex", the answer would be switched.