Calculus Solutions Ex#2.2

 

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Q.07
Solution:
Given $$ \begin{aligned} & m=f^{\prime}(3)=5 \\ \\ & x_{1}=3, \quad y_{1}=f(3)=-1 \end{aligned} $$ Equation of tangent line $$ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ \\ y-(-1) & =5(x-3) \\ \\ y+1 & =5(x-3) \\ \\ y & +1=5 x-15 \\ \\ y & =5 x-15-1 \\ \\ y & =5 x-16 \end{aligned} $$
Q.08 $f(-2)=3, f^{\prime}(-2)=-4$, find an equation for the tangent line to the graph of $y=f(x)$ at $x=-2$.
Solution:
$$ & $$ Given $$\quad m=f^{\prime}(-2)=-4$$ $$ x_{1}=-2, \quad y_{1}=f(-2)=3 $$ Equation of tangent line $$ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ \\ y-(3) & =-4(x-(-2)) \\ \\ y-3 & =-4(x+2) \\ \\ y-3 & =-4 x-8 \\ \\ y & =-4 x-8+3 \\ \\ y & =-4 x-5 \end{aligned}$$
Q.09
Use definition 2.2 .1 to find $f^{\prime}(x)$ and then find the tangent line to the graph of $y=f(x)$ at $x=a$.
Solution:
$$\quad f(x)=2 x^{2} ; \quad a=1$$ Definition 2.1 .1 $$ \begin{aligned} & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2(x+h)^{2}-2 x^{2}}{h} \\ \\ & \therefore f(x)=2 x^{2} \Rightarrow f(x+h)=2(x+h)^{2} \\ \\ & f^{\prime}(x)= \lim _{h \rightarrow 0} \frac{2\left(x^{2}+h^{2}+2 x h\right)-2 x^{2}}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 x^{2}+2 h^{2}+4 x h-2 x^{2}}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 h^{2}+4 x h}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h(2 h+4x)}{h} \\ \\ & =\lim_{h\rightarrow 0}(2h+4x)=4 x \\ \\ & f^{\prime}(x)=4 x \\ \\ & m=f^{\prime}(a)=4(1)=4 \\ \\ & \therefore \quad a=1 \\ \\ & x_{1}=a=1, \quad y_{1}=f(a)=2(1)^{2}=2 \end{aligned} $$ Equation of tangent line $$ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ \\ y-2=4(x-1) \Rightarrow & y-2=4 x-4 \\ \\ & y=4 x-4+2 \\ \\ y & =4 x-2 \end{aligned} $$
Q.10
Solution:
$$ \begin{aligned} & f(x)=1 / x^{2} ; \quad a=-1 \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^{2}}-1 / x^{2}}{h} \\ \\ & \therefore f(x)=1 / x^{2} \Rightarrow  f(x+h)=\frac{1}{(x+h)^{2}} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{2}-\left(x^{2}+h^{2}+2 x h\right)}{h x^{2}(x+h)^{2}} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{2}-x^{2}-h^{2}-2 x h}{h x^{2}\left(x^{2}+h^{2}+2 x h\right)} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-h^{2}-2 x h}{h x^{2}\left(x^{2}+h^{2}+2 x h\right)} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h\left(-h-2 x\right)}{h x^{2}\left(x^{2}+h^{2}+2 x h\right)} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-h-2 x}{x^{2}\left(x^{2}+h^{2}+2 x h\right)} \\ \\ & f^{\prime}(x)=\frac{0-2 x}{x^{2}\left(x^{2}+0+0\right)}=-\frac{2 x}{x^{4}} \end{aligned} $$  $$ f^{\prime}(x)=-2 / x^{3} $$  put $$x_{1}=a=-1$$ $$ \begin{aligned} \\ \\ & m=f^{\prime}(a)=-2 /(-1)^{3}=2 \\ \\ \text {and} \quad & y_{1}=f(-1)=1 \end{aligned} $$ Equation of tangent line $$ \begin{gathered} y-y_{1}=m\left(x-x_{1}\right) \\ \\ y-1=2(x-(-1)) \\  \\ y-1=2(x+1) \\ \\ y-1=2 x+2 \\  \\ y=2 x+2+1 \\ \\ y=2 x+3 \end{gathered} $$
Q.11
Solution:
$$ \begin{aligned} & f(x)=x^{3} ; a=0 \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h} \\ \\ & \because f(x)=x^{3} \Rightarrow f(x+h)=(x+h)^{3} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}- x^{3}}{h} \\ \\ \quad \therefore\left(a+b\right)^{3} & =a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h^{3}+3{x }^{2} h+3 x h^{2}}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+3 x^{2}+3 x h\right)}{h} \\ \\ & f^{\prime}(x)=\lim _{h \rightarrow 0}\left(h^{2}+3 x^{2}+3 x h\right) \\ \\ & f^{\prime}(x)=0+3 x^{2}+0=3 x^{2} \\ \\ & f^{\prime}(x)=3 x^{2} \\ \\ & \text { Put } x=a=0 \\ \\ & x_{1}=a=0, \\ \\ & m=f^{\prime}(a)=0,\quad y_{1}=f(a)=0 \end{aligned} $$ Equation of tangent line $$ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ \\ y-0 & =0(x-0) \\ \\ y & =0 \end{aligned} $$