From the question, we can deduce that at first it would be better to have g not as a function of x and m, but as a function of f. We have expressions of x and m as functions of f. Hence, we can substitute these expressions into g(x,m) and get the desired form of g: g(f)=(−f2+5f)2+2(f−1)+240=f2(5−f)2+2f−2+240=f2(25−10f+f2)+2f+238=f4−10f3+25f2+2f+238.
Next, we take "very small" in the question as "infinitesimal". Now, we know that we deal with differentials. In our case, an infinitesimal positive change in the supply of food leads to an infinitesimal change in probability of winning, i.e. +df or just df causes dg, and we are looking for the very dg. Then, dg relates to df by the formula dg=g′(f)df, where g′(f) is derivative of g with respect to f. We can obtain this derivative by the well-known, simple derivative rules for polynomials: g′(f)=4f3−10∗3f2+25∗2f+2∗1+0=4f3−30f2+50f+2.
By the problem condition, f=1.5. Then, finally, we can compute dg at f=1.5:
dg=g′(1.5)df=(4(1.5)3−30(1.5)2+50(1.5)+2)df=23df.
So, the answer is dg=23df.
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