(ll'J points) Recall from calculus that given some flmction 53(3),, the a: you get from solving Ed? = I} is called a critical point ofy this means it could be a minimiser or a masimiser for g. In this question} we will explore some basic properties and build some intuition on why} for certain lom functions such as the MSE loss} the critical point of the lom wl always be the minimiser of the loss. IGiven some linear model at) = "(I for some real scalar of, we can write the the mean squared error (LEE) loss of the model f given the observed data {rhyihi = 1} . . . in. as gs. arr)\"- (a) (1 point) Let's break the loss function above into individual terms. Complete the following sentence by filling in the blanks using one of the options in the parenthesis following each of the blanks: The MSE loss function can be viewed as a sum of n _---- (linear/quadratic/log- arithmic/exponential) terms, each of which can be treated as a function of (I; /y:/ 7). (b) (3 points) Let's investigate one of the n functions in the summation in the MSE loss function. Define g;(y) = (y, - yr;)' for i = 1, ..., n. Recall from calculus that we can use the 2nd derivative of a function to describe its curvature about a certain point (if it is facing concave up, down, or possibly a point of inflection). You can take the following as a fact: A function is convex if and only if the function's 2nd derivative is non-negative on its domain. Based on this property, verify that g; is a convex function