Problems

Solution:

1. \(y = e^x(2x^2-5x+2) = e^x(2x-1)(x-2)\), we also have \(y' = e^x(2x^2-5x+2 + 4x-5) = e^x(2x^2-x-3) = e^x(2x-3)(x+1)\) \(y(-1) = \frac{9}{e}\), \(y(\frac32) = -e^{3/2}\)
   

   + - ↺
   If \(k < -e^{3/2}\) there are no solutions. If \(k = -e^{3/2}\) there is a unique solution. If \(-e^{3/2} < k \leq 0\) there are two solutions. If \(0 < k < \frac{9}{e}\) there are three solutions. Otherwise there is a unique solution.
2. 

   + - ↺

Solution:

1. \(\,\)
   

   + - ↺
   \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
2. There are no solutions if \(a < c < b\):
   

   + - ↺
   There is one solution if \(a=c\) or \(a = b\)
   

   + - ↺
   And there are two solution if \(c \not \in [a,b]\)
   

   + - ↺
   There is exactly one solution unless....
   

   + - ↺
   ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
   

   + - ↺

Solution:

1. \(\,\) \begin{align*} && x &= (a-x)^{\frac12} \\ \Rightarrow && x^2 &= a - x \\ \Rightarrow && 0 &= x^2 + x - a \end{align*} This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise. \begin{align*} && x & = \left ( (1+a)x - a \right)^{\frac13} \\ \Leftrightarrow && x^ 3 &= (1+a)x - a \\ \Leftrightarrow && 0 &= x^3- (1+a)x + a \\ \Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\ \end{align*} Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore: One solution if \(a < -\frac14\) Two solutions if \(a = -\frac14, 2\) Three solutions if \(a > -\frac14, a \neq 2\)
2. \(\,\) \begin{align*} && x &= (b+x)^{\frac12} \\ \Rightarrow && x^2 &= b + x \\ \Rightarrow && 0 &= x^2 - x - b \end{align*} This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\). Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)

Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

+ - ↺

1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}