MTH 2610: Final Practice 1

We will first evaluate the limit by multiplying both numerator and the denominator by \( \sqrt{ 7x+9 }+4 \). The limit becomes: \begin{eqnarray*} \lim_{x\to 1}\frac{x-1}{\sqrt{ 7x+9 }-4}&=&\lim_{x\to 1}\frac{\left(x-1\right)\cdot \left(\sqrt{ 7x+9 }+4\right)}{ 7x+9 -16}=\lim_{x\to 1}\frac{\left(x-1\right)\cdot \left(\sqrt{ 7x+9 }+4\right)}{ 7(x-1)}\newline &=&\lim_{x\to 1}\frac{\sqrt{ 7x+9 }+4}{7}=\frac{8}{7}.\end{eqnarray*}

The given equation is equivalent to \( x^2-3x-54={2}^{4 } \). Its solutions are \( x_1=-7 \) and \( x_2=10 \).

The function \( y \) intersects the \( x \) axis at two points \( 0 \) and \( 4 \). Therefore the required area is \[ A=\int_0^{4}\left(48x-12x^2\right)\,dx=128.\]

We will use that \( df=f^{\prime}(x)\,dx \) and \( \Delta f=f(x)-f(x_0) \). Using the approximation \( df\approx \Delta f \) we obtain \[ f(x)\approx f(x_0)+f^{\prime}(x_0)\cdot (x-x_0)=f(x_0)+ 6 \cdot(x-x_0) =10.8.\]

We first solve the equation \( y=\frac{4x-2}{ x+3} \). We obtain \( yx+3y=4x-2 \) which gives us \( x(y-4)=-3y-2 \) which implies that \( x=-\frac{3y+2}{y-4} \). Therefore \( \displaystyle f^{-1}(x)=\frac{3x+2}{4-x} \).

Using the implicit differentiation we obtain: \[ 6x^{5}e^y+x^{6}e^yy^{\prime}=y^{3}+3xy^{2}y^{\prime}.\] Solving for \( y’ \) gives \[ 6x^{5}e^y-y^{3}=y^{\prime}\left(3xy^2-x^{6}e^y\right).\]

Therefore \[ y^{\prime}=\frac{6x^{5}e^y-y^{3} }{ 3xy^2-x^{6}e^y }.\]

We will use the fact that \( y=\frac1{2}\ln\left(x-8\right)+\ln(x+3)-\ln x \). The derivative is now easy to calculate and the correct result is \[y^{\prime}=\frac1{2(x-8)}+\frac1{x+3}-\frac1x.\]

Using \( \ln\left(x^{8}\right)=8\ln x \), we first simplify the function \( f \) in the following way: \( f(x)=x^{8\ln x} \). We now use that \( f(x)=e^{\ln f(x)} \) to obtain \( \displaystyle f(x)=e^{\ln \left(x^{8\ln x} \right)}=e^{8\ln x\cdot \ln x}=e^{8\left(\ln x\right)^2} \). We now have \[ f^{\prime}(x)=\left(e^{8\left(\ln x\right)^2}\right)^{\prime}= e^{8\left(\ln x\right)^2}\cdot \left(8\left(\ln x\right)^2\right)^{\prime}=8 e^{8\left(\ln x\right)^2}\cdot 2\ln x\cdot \frac1x= \frac{16\ln x\cdot e^{8\left(\ln x\right)^2}}x. \]

From \( xy=12 \) we conclude that \( y=\frac{12}x \) and \( 3x+4y=3x+\frac{48}x \). Let \( f(x)=3x+\frac{48}x \). We need to find the minimum of \( f \). We will first find the critical points of \( f \). There are three categories of critical points. The first category consists of the endpoints of the interval on which \( f \) is defined. The only endpoint is \( x=0 \) and the function is not defined for it while \( \lim_{x\to 0}f(x)=+\infty \). The second category consists of those values of \( x \) for which \( f \) is not defined. The derivative does not exist at \( 0 \) and we already considered this value. The third category consist of those values of \( x \) for which \( f^{\prime}(x)=0 \). The only positive solution to \( f^{\prime}(x)=0 \) is \( x=4 \). Then we have \( f(4)=24 \).

The function is increasing for \( x< 5 \) and decreasing for \( x> 5 \). Thus \( f \) attains a maximum at \( 5 \). The function is concave up except at \( 5 \) where it is not differentiable.

Let us denote by \( y \) the length of the edge of the rectangle parallel to \( CA \) and by \( x \) the length of the edge parallel to \( AB \). Let us denote by \( \beta \) the angle of \( \triangle ABC \) corresponding to the vertex \( B \). Let us denote by \( Y \) the vertex of the rectangle that belongs to \( AB \). Then we have \( \tan\beta=\frac{6}{12}=\frac{PY}{YB}=\frac{y}{12-x} \). Therefore \( y=\frac{3}{6}\cdot \left(12-x\right) \). The area of the rectangle is \( A= x\cdot y=x\cdot \frac{3}{6}\cdot \left(12-x\right) \). We now need to find the maximum of the function \( f(x)=x\cdot \frac{3}{6}\cdot \left(12-x\right) \). We will first find the critical values of the function. There are three categories of critical values. The first category consists of the endpoints. The endpoints for the interval for \( x \) are \( 0 \) and \( 12 \). The second category consists of those values of \( x \) for which \( f^{\prime} \) is not defined. There are no such values. The third category consists of those values for \( x \) for which \( f^{\prime}(x)=0 \). We have that \( f^{\prime}(x)=\frac{3}{6}\cdot \left(12-2x\right) \). The only solution to \( f^{\prime}(x)=0 \) is \( x=6 \).

Therefore the critical values are \( x=0 \), \( x=12 \) and \( x=6 \). We have \( f(0)=0 \), \( f(12)=0 \), and \( f(6)=18 \). Thus the maximal area is \( 18 \).

The given integral can be written as the sum of two integrals: \[ \int_{4}^{6}\frac{2\sqrt{x-1} +4}{x\sqrt{x-1} }\,dx=\int_{4}^{6}\frac{2}x\,dx+4\int_{4}^{6}\frac{dx}{x\sqrt{x-1}}.\] The first integral is easy to evaluate: \( \displaystyle\int_{4}^{6}\frac{2}{x}\,dx= 2\ln6-2 \ln4 \). In the second integral we use the substitution \( u=\sqrt{x-1} \). Then we get \( x=u^2+1 \) and \( dx=2u\,du \). The bounds of integration become \( \sqrt{3}\leq u\leq \sqrt{5} \). Therefore \[ \int_{4}^{6}\frac{ 4}{x\sqrt{x-1} }\,dx=4\int_{\sqrt{3}}^{\sqrt{5}}\frac{2du}{u^2+1}=8\arctan\sqrt{5}-8\arctan\sqrt{3}.\] Thus \[ \int_{4}^{6}\frac{2\sqrt{x-1} +4}{x\sqrt{x-1} }\,dx= 2\ln6-2 \ln4+8\arctan\sqrt{5}-8\arctan\sqrt{3}.\]

• (a) We first need to find the intersection points of the graphs of the functions \( y=f(x) \) and \( y=g(x) \). The solutions to the equation \( x^3=16x-6x^2 \) that satisfy \( x\geq 0 \) are \( x=0 \) and \( x=2 \).
  

• (b) The required area is \[ A=\int_0^{2}g(x)\,dx-\int_0^{2}f(x)\,dx= \left.\left(\frac{16x^2}2-\frac{6x^3}3\right)\right|_{x=0}^{x=2}-\left.\frac{1}{4}\cdot x^{4}\right|_{x=0}^{x=2}=\frac{144}{12} =12.\]

Yes such two functions do exist. One example is \( f(x)=-12x \) and \( g(x)=x \). Then \( h(x)=0 \) which is clearly not onto.

Consider the function \( \displaystyle F(x)=\int_0^x\left(f(t)-e^{-t^4}\right)\,dt \). Since \( \displaystyle\int_0^{ 7 }f(t)\,dt=\int_0^{7}e^{-t^4}\,dt \) we conclude that \( F(7)=0 \). Since \( F(0)=0 \) we can use Rolle‘s theorem to conclude that there exists \( c\in[0,7] \) such that \( F^{\prime}(c)=0 \). Using the Fundamental Theorem of Calculus we conclude that \( F^{\prime}(x)=f(x)-e^{-x^4} \). Therefore there exists \( c\in[0,7] \) such that \( f(c)-e^{-c^4}=0 \) which is equivalent to \( e^{c^4}f(c)=1 \).