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  • Fill in the blanks in each of the following The curves y = 4x^2 + 2x - 8 and y = x^3 - x + 13 touch each other at the point_____.

Fill in the blanks in each of the following
The curves y = 4x^2 + 2x - 8 and y = x^3 - x + 13 touch each other at the point_____.

Answers (1)

Given the first curve is y = 4x^2 + 2x - 8

Applying first derivative and get

\frac{d y}{d x}=\frac{d\left(4 x^{2}+2 x-8\right)}{d x}
Apply sum rule and 0 is the differentiation of the constant term is 0,so
\frac{d y}{d x}=4 \frac{d\left(x^{2}\right)}{d x}+2 \frac{d(x)}{d x}+0
Apply power rule and get
\frac{d y}{d x}=8 x+2
This is the slope of the first curve; let m_{1}  be equal to this.
\Rightarrow m_{1}=8 x+2 \ldots \ldots(0)
The second curve is y=x^{3}-x+13
Applying first derivative and get
\frac{d y}{d x}=\frac{d\left(x^{3}-x+13\right)}{d x}

Apply sum rule and 0 is the differentiation of the constant term, so

\begin{aligned} &\frac{d y}{d x}=\frac{d\left(x^{3}\right)}{d x}-\frac{d(x)}{d x}+0\\ &\text { Apply power rule and get }\\ &\frac{d y}{d x}=3 x^{2}-1 \end{aligned}

This is the slope of the second curve; let m_2 be equal to this.

\Rightarrow m\textsubscript{2}=3x\textsuperscript{2}-1 \ldots \ldots (ii)

Now the slopes must be equal, because they touch each other, i.e.,

m\textsubscript{1}=m\textsubscript{2}

\\ { \Rightarrow 8x+2=3x\textsuperscript{2}-1}\\ { \Rightarrow 3x\textsuperscript{2}-8x-1-2=0}\\ { \Rightarrow 3x\textsuperscript{2}-8x-3=0}\\

Split middle term and get

\\ { \Rightarrow 3x\textsuperscript{2}+x-9x -3=0}\\ { \Rightarrow x(3x+1)-3(3x+1)=0}\\ { \Rightarrow (3x+1)(x-3)=0}\\ { \Rightarrow (3x+1)=0 or (x-3)=0}\\

\\ \Rightarrow x=-\frac{1}{3} or x=3 \\
Substitute x=-\frac{1}{3} in both the equations and get
For first curve, y=4 x^{2}+2 x-8
\\ \Rightarrow y=4\left(-\frac{1}{3}\right)^{2}+2\left(-\frac{1}{3}\right)-8$ \\$\Rightarrow y=4\left(\frac{1}{9}\right)-\frac{2}{3}-8$ \\$\Rightarrow \mathrm{y}=\frac{4-2 \times 3-8 \times 9}{9}=-\frac{74}{9}
For second curve, y=x^{3}-x+13
\Rightarrow y=\left(-\frac{1}{3}\right)^{3}-\left(-\frac{1}{3}\right)+13

\begin{aligned} &\Rightarrow \mathrm{y}=-\frac{1}{27}+\frac{1}{3}+13\\ &\Rightarrow \mathrm{y}=\frac{-1+1 \times 9+13 \times 27}{27}=\frac{8}{27}+13\\ &\text { Thus at }\\ &\mathrm{x}=-\frac{1}{3^{\prime}} \text { curves don't touch } \end{aligned}

Substitute x=3 in both equations

For first curve, y=4 x^{2}+2 x-8

\\ { \Rightarrow y=4(3)\textsuperscript{2}+2(3)-8}\\ { \Rightarrow y=4(9)+6-8=34}\\

For second curve, y=x^{3}-x+13

\\ { \Rightarrow y= (3)\textsuperscript{3} - (3) + 13}\\ { \Rightarrow y=27-3+13=37}\\

Hence at x=3 both curves don't touch

So, the curves y = 4x\textsuperscript{2} + 2x - 8  and  y = x\textsuperscript{3} - x + 13 do not touch each other.
 

Posted by

infoexpert22

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