MTH 2610: Midterm 1 Practice 1

We first evaluate \( g(9)=11\cdot 9-18=81 \). Therefore \( f(g(9))=f(81)=9 \).

The limit can be calculated by dividing both numerator and denominator by \( x^2 \). The limit becomes \[ \lim_{x\to 0}\frac{12x^3+5x^2+15x^4}{x^4+x^3+x^2}=\lim_{x\to 0}\frac{ 12x+5+15x^2}{x^2+x+1}=\frac{\lim_{x\to 0}\left(12x+5+15x^2\right)}{\lim_{x\to 0}\left(x^2+x+1\right)}=5.\]

• (a) Notice that the numerator can be factorized in the following way \( x^2-4x-21=(x-7)(x+3) \). Therefore the limit is equal to: \[ \lim_{x\to 7} \frac{x^2-4x-21}{x-7}=\lim_{x\to 7} \frac{(x-7)(x+3)}{x-7}=\lim_{x\to7}(x+3)=10.\]

• (b) We will multiply both the numerator and the denominator by \( \displaystyle \sqrt{11+9x^2}+\sqrt{11} \). The limit transforms into the following one \begin{eqnarray*}\lim_{x\to 0}\frac{x^2}{\sqrt{11+9x^2}-\sqrt{11}}&=&\lim_{x\to 0}\frac{x^2\left(\sqrt{11+9x^2}+\sqrt{11}\right)}{ 11+9x^2- 11}\newline &=&\frac1{9}\lim_{x\to 0}\left( \sqrt{11+9x^2}+\sqrt{11} \right)=\frac{2\sqrt{11}}{9}.\end{eqnarray*}

We first need to modify the given expression into a form that involves only one fraction. This can be done by first finding the common denominator of the fractions \( \displaystyle \frac1{1+4x} \) and \( \displaystyle \frac1{1+x^2} \). The required limit becomes: \begin{eqnarray*}\lim_{x\to 0}\frac{\frac1{1+4x}-\frac1{1+x^2}}{ 7x}&=& \lim_{x\to 0}\frac{\frac{1+x^2-1-4x}{\left(1+4x\right)\left(1+x^2\right)}}{7x}=\lim_{x\to 0}\frac{x^2-4x}{7x\left(1+4x\right)\left(1+x^2\right)}\newline &=& \lim_{x\to 0}\frac{x\left(x-4\right)}{7x\left(1+4x\right)\left(1+x^2\right)}=\lim_{x\to 0}\frac{x-4}{7\left(1+4x\right)\left(1+x^2\right)}\newline &=&-\frac{4}{7}. \end{eqnarray*}

• (a) The derivative of \( f \) satisfies \( f^{\prime}(x)=16x-5 \) and \( f^{\prime}(1)=11 \).

• (b) The slope of the graph of the function at \( (1,2) \) is \( f^{\prime}(1)=11 \). The equation of the tangent line is \( y-f(1)=f^{\prime}(1)\cdot(x-1) \), i.e. \( y=11x-9 \).

• (a) We will use the product rule to solve this problem. \begin{eqnarray*}f^{\prime}(x)&=&\left(x+7\sin x\right)^{\prime}\cdot\left(\cos x-3\right)+\left(x+7\sin x\right)\cdot\left(\cos x-3\right)^{\prime}\newline &=& \left(1+7\cos x\right)\left(\cos x-3\right)+\left(x+7\sin x\right)\cdot (-\sin x)\newline &=&7\left(\cos^2x-\sin^2x\right)-20\cos x-x\sin x-3.\end{eqnarray*}

• (b) We will use the quotient rule to solve this problem. \begin{eqnarray*} f^{\prime}(x)&=&\frac{\left(x-7\right)^{\prime}\left(x^2+3\right)-\left(x-7\right)\left(x^2+3\right)^{\prime}}{\left(x^2+3\right)^2}\newline &=&\frac{ x^2+3 -2x\left(x-7\right)}{\left(x^2+3\right)^2}\newline &=&\frac{- x^2+14x+3}{\left(x^2+3\right)^2} \end{eqnarray*}

• (c) Before applying the quotient rule we can simplify the numerator using the distribution property \( (x+7)(x-3)=x^2+4x-21 \). Therefore \begin{eqnarray*}f^{\prime}(x)&=&\frac{\left(2x+4\right)\left(x^2+\sin x\right)-\left(x^2+4x-21\right)\left(2x+\cos x\right)}{\left(x^2+\sin x\right)^2}\newline &=&\frac{-4x^2+42x+2x\sin x+4\sin x-x^2\cos x-4x\cos x+21\cos x}{\left(x^2+\sin x\right)^2}. \end{eqnarray*}

The function \( f \) is a constant function. Therefore, there is a real number \( c \) such that \( f(x)=c \) for all \( x \). In particular, that means that \( f(g(x))=f\left(x^2\right)=f(x^3)=\cdots=c \), and \( g(f(x))=g(c) \). Only function given in C is not necessarily a constant function, thus that is the only function that can have a nonzero slope.