GRE Problems - Variants

Solution This problem is essentially the same as the previous one [GRE 1268, #30a], except that we need to work out the solutions in a few different cases. Define \(f(x)=3x^5-20x^3+45x\). Then saying that the equation \(3x^5+45x=20x^3+c\) has \(k\) solutions is the same as saying that \(f(x)-c\) has \(k\) \(x\)-intercepts. So we study the shape of the graph of \(f(x)\). The first derivative \(f'(x)= 15 x^4 -60 x^2+45\) factors as \(15(x-1)(x+1)(x-\sqrt{3})(x+\sqrt{3})\). From a sign chart for the derivative, we find that \((-\sqrt{3},-12\sqrt{3})\) and \((1,28)\) are points of local maximum, and \((-1,-28)\) and \((\sqrt{3},12 \sqrt{3})\) are points of local minimum. Also, \(f(x)\) is a \(5\)-th degree polyomial with positive leading coefficient. So it will go to \(-\infty \) as \(x\rightarrow -\infty\), and to \(+\infty\) as \(x\rightarrow \infty\). We are trying to find the values of \(c\) so that \(f(x) -c\) has a certain number of \(x\)-intercepts. So we are interested in the shifted function \(f(x)-c\), whose points of local maximum and minimum will be just the same, but shifted vertically by \(-c\). So the shape of the graph is as follows: \(\img{1268_30b.png}{0em}{}{12em}\) Choosing a specific value for \(c\) is the same thing as deciding where to draw the \(x\)-axis on the above graph. For example, drawing the \(x\)-axis through the local minimum \((-1,-28-c)\) is the same thing as saying that \(-28-c=0\), or \(c=-28\). So we can find the number of \(x\)-intercepts by drawing the \(x\)-axis in various positions. For example, if we draw the \(x\) axis below the local minimum at \((-1,-28-c)\), as in the picture below, \(\img{1268_30b1.png}{0em}{}{12em}\) there is only one \(x\)-intercept. This means that all corresponding values of \(c\) will belong to the set \(S_1\). But drawing the \(x\)-axis that way means that \(-28-c\gt 0\), or \(c\lt -28\). This means that all values of \(c\) less than \(-28\) will be in \(S_1\), or, in interval notation, \((-\infty, -28) \) is a subset of \(S_1\). But this is not all of \(S_1\), because if we draw the \(x\)-axis as shown: \(\img{1268_30b2.png}{0em}{}{12em}\) then there will be only one \(x\)-intercept. The last picture says that \(-12\sqrt{3}-c\lt 0\) and \(12\sqrt{3} -c\gt 0\). Solving the two inequalities for \(c\), we find \(-12\sqrt{3} \lt c \lt 12\sqrt{3}\). This means that the interval \((-12\sqrt{3},12\sqrt{3})\) is also a subset of \(S_1\). And there is one more way to get a single \(x\)-intercept: draw the \(x\)-axis as shown: \(\img{1268_30b3.png}{0em}{}{12em}\) This corresponds to \(28-c\lt 0\), or \(c\gt 28\), so that the interval \((28,\infty)\) is also a subset of \(S_1\). It should be clear looking at the graph that there are no other ways to get exactly one \(x\)-intercept. So we have found the set \(S_1\): \[S_1=(-\infty,-28)\cup (-12\sqrt{3},12\sqrt{3})\cup (28,\infty).\] It is also clear from the graph that it is impossible for the graph to have no \(x\)-intercepts. This means that \(S_0\) is the empty set. To get two \(x\)-intercepts, we can draw the \(x\)-axis in the following four ways: \[\img{1268_30b4.png}{0em}{}{10em}, \img{1268_30b5.png}{0em}{}{10em}\] \[ \img{1268_30b6.png}{0em}{}{10em}, \img{1268_30b7.png}{0em}{}{10em}\] Each way corresponds to a specific value of \(c\), obtained by setting the \(y\)-coordinate of the \(x\)-intercept to zero, and so the set \(S_2\) has exactly four points: \[S_2=\{-28,-12\sqrt{3},12\sqrt{3},28\}.\] There are two ways to draw the \(x\)-axis and get three \(x\)-intercepts: \[\img{1268_30b8.png}{0em}{}{12em}, \img{1268_30b9.png}{0em}{}{12em}\] These two ways correspond to the inequalities \(12\sqrt{3}-c\lt 0\), \(28-c\gt 0\), that means \(12\sqrt{3}\lt c\lt 28\), and \(-28-c\lt 0\), \(-12\sqrt{3}-c \gt 0\), that is \(-28\lt c\lt -12\sqrt{3}\). This gives us \(S_3\): \[S_3=(-28,-12\sqrt{3})\cup(12\sqrt{3},28).\] It's impossible for the graph to have \(4\) or more \(x\)-intercepts. So \(S_k=\emptyset\) for \(k\geq 4\).