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Chapter 5: Problem 76

\(y=x^{3}-x^{2}-x+2\) \(\frac{d y}{d x}=3 x^{2}-2 x-1=0\) at \(x=1\) eqn of tangent \(y=1\) solving with curve, \(x^{3}-x^{2}-x+1=0\) \((x-1)\left(x^{2}-1\right)=0\) \(\mathrm{x}=\pm \mathrm{l}\)

Short Answer

Expert verified
#Answer# The tangent line's equation is \(y=1\). The intersection points are at \(x=-1, 0,\) and \(1\).

Step by step solution

01

Compute the derivative of the function

To find the equation of the tangent line, we first need to find the derivative of the function. The derivative of the function is given by: \(\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}-x^{2}-x+2\right)\) Using the power rule, we find the derivative to be: \(\frac{d y}{d x}=3x^{2}-2x-1\)
02

Evaluate the derivative at the given point

We are given that the tangent touches the curve at \(x=1\). So, let's find the tangent's slope at this point by plugging the value into the derivative: \(\frac{d y}{d x}(1)=3(1)^{2}-2(1)-1=3-2-1=0\)
03

Find the equation of the tangent line

Since the derivative is zero at \(x=1\) and we know that the tangent line should have the same slope as the curve at this point, the tangent line should be horizontal. Therefore, its equation will just be a constant value, which is equal to the function's value at \(x=1\): \(y(1)=1^{3}-1^{2}-1+2=1-1-1+2=1\) Thus, the equation of the tangent line is \(y=1\).
04

Determine the intersection points of the tangent line with the curve

To find the points where the tangent line intersects the curve, we need to set the two equations equal: \(x^{3}-x^{2}-x+2=1\) This simplifies to the following equation: \(x^{3}-x^{2}-x+1=0\) Factoring this equation, we get: \((x-1)(x^{2}-1)=0\) Which yields the solutions \(x=-1, 0,\) and \(1\). Since we already knew about the intersection point at \(x=1\), the other two intersection points are \(x=-1\) and \(x=0\).

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Most popular questions from this chapter

\(x=t^{2}+t+1\) \(\frac{d x}{d t}=2 t+1\) \(y=t^{2}-t+1\) \(\frac{d y}{d t}=2 t-1\) \(\frac{d y}{d x}=\frac{2 t-1}{2 t+1}\) $\Rightarrow y-\left(t^{2}-t+1\right)=\frac{2 t-1}{2 t+1}\left(x-\left(t^{2}+t+1\right)\right)$ $\Rightarrow\left(t-t^{2}\right)=\left(\frac{2 t-1}{2 t+1}\right)\left(-\left(t^{2}+t\right)\right)$ \(\Rightarrow 2 t^{2}+t-2 t^{3}-t^{2}=t^{2}+t-2 t^{3}-2 t^{2}\) \(\Rightarrow 2 t^{2}=0\) \(\Rightarrow t=0\)

A) As \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R\) \(\Rightarrow a=2 R \sin A\) \(\Rightarrow \frac{d a}{d A}=2 R \cos A\) \(\Rightarrow \frac{d a}{\cos A}=2 R d A\) Similarly, \(\frac{d b}{\cos B}=2 R d B, \frac{d c}{\cos c}=2 R d C\) \(\quad \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}+1=\) \(2 R(d A+d B+d C)+1\) \(\quad A s A+B+C=\pi\) \(\Rightarrow d A+d B+d C=0\) using eq (I) \(|\mathbf{m}|=1\) B) $\begin{aligned} & \Rightarrow \mathrm{m}=\pm 1 \\ \mathrm{x}^{2} \mathrm{y}^{2}=& 16 \end{aligned}$ \(2 x^{2} y \frac{d y}{d x}+2 x y^{2}=0\) \(\frac{d y}{d x}=-\frac{y}{x}=1\) Subtangent \(=y / d y / d x=2=|\mathrm{k}|\) \(\Rightarrow \mathrm{k}=\pm 2\) C) \(y=2 e^{2 x}\) \(\frac{d y}{d x}=4 e^{2 x}=4\) for yaxis $\tan ^{-1} 4=\frac{\pi}{2}-\cot ^{-1}\left(\frac{8 n-4}{3}\right)=\tan ^{-1}\left(\frac{8 n+4}{3}\right)$ \(12=|8 n-4|\) \(\mathrm{n}=2\) or \(-1\) D) \(\mathrm{x}=\underline{\mathrm{e}^{\sin y}}\) \(1=e^{\sin y} \cos y \frac{d y}{d x}\) At \((1,0) \Rightarrow \frac{d y}{d x}=1\) eqn of Normal \(\rightarrow y=-(x-1)\) \(y+x-1=0\) Area of \(\Delta=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\) \(\Rightarrow|2 t+1|=3\) \(2 t+1=\pm 3 \Rightarrow t=1\) or \(-2\) $\mathrm{A} \rightarrow(\mathrm{PQ}), \mathrm{B} \rightarrow(\mathrm{RS}), \mathrm{C} \rightarrow(\mathrm{RQ}), \mathrm{D} \rightarrow(\mathrm{PS})$

$\mathrm{S}=30\left(1-\mathrm{e}^{(\ln 5 / 6) 5}\right)=30\left(1-\left(\frac{5}{6}\right)^{5}\right)$

\(y=\left(\frac{x}{2}-a\right)^{2}+a-2\) \(4(y-(a-2))=(x-2 a)^{2}\) Vertex \(\Rightarrow \mathrm{h}=2 \mathrm{a}, \mathrm{k}=\mathrm{a}-2\) Locus of vertex \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}-2\) \(\Rightarrow 2 y=x-4\)

For \(y=x(c-x)\) \(\frac{d y}{d x}=c-2 x\) For \(y=x^{2}+a x+b\) \(\frac{d y}{d x}=2 x+a\) At \((1,0)\) \(0=(C-1) \Rightarrow C=1\) \(0=1+a+b, 1-2=2+a\) \(\Rightarrow \quad a=-3\) \(\Rightarrow b=2\)
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