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Chapter 2: Q7E (page 77)

5-8 Find an equation of the tangent line to the curve at the given point.

7. \(y = \frac{{x + 2}}{{x - 3}},\,\,\left( {2, - 4} \right)\)

Short Answer

Expert verified

The equation for the tangent line to the curve at the given point is \(y = - 5x + 6\).

Step by step solution

01

Tangent line of a function

For a curve \(y = f\left( x \right)\) a tangent lineat a particular point is a line through that

point with a slope. For example let the point be \(P\left( {b,f\left( b \right)} \right)\) then tangent line at this point will be line through this point with slope :

\(m = \mathop {\lim }\limits_{x \to b} \frac{{f\left( x \right) - f\left( b \right)}}{{x - b}}\)

This limit should be exist.

02

Finding the equation of a tangent line of the given curve at the given point

The curve is given by \(y = \frac{{x + 2}}{{x - 3}}\).

Now let \(f\left( x \right) = \frac{{x + 2}}{{x - 3}}\) and the given point is \(\left( {2, - 4} \right)\).

Thus slope of the tangent line will be:

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}}\\ &= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{x + 2}}{{x - 3}} - \left( { - 4} \right)}}{{x - 2}}\\ &= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{x + 2}}{{x - 3}} + 4}}{{x - 2}}\end{aligned}\)

Further solving we get,

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{x + 2 + 4x - 12}}{{x - 3}}}}{{x - 2}}\\ &= \mathop {\lim }\limits_{x \to 2} \frac{{5x - 10}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\ &= \mathop {\lim }\limits_{x \to 2} \frac{{5\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\ &= \mathop {\lim }\limits_{x \to 2} \frac{5}{{\left( {x - 3} \right)}}\\ &= - 5\end{aligned}\)

Hence slope of the tangent lineis \( - 5\).

Now the equation of the tangent linethrough the point \(\left( {2, - 4} \right)\) will be:

\(\begin{aligned}y - \left( { - 4} \right) &= m\left( {x - 2} \right)\\y + 4 &= - 5\left( {x - 2} \right)\\y &= - 5x + 6\end{aligned}\)

Hence the equation of the tangent line at point \(\left( {2, - 4} \right)\) is \(y = - 5x + 6\).

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Most popular questions from this chapter

Suppose f and g are continuous functions such that \(g\left( {\bf{2}} \right) = {\bf{6}}\) and \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{2}}} \left( {{\bf{3}}f\left( x \right) + f\left( x \right)g\left( x \right)} \right) = {\bf{36}}\). Fine \(f\left( {\bf{2}} \right)\).

For the function g whose graph is shown, find a number a that satisfies the given description.

(a)\(\mathop {{\bf{lim}}}\limits_{x \to a} g\left( x \right)\) does not exist but \(g\left( a \right)\) is defined.

(b)\(\mathop {{\bf{lim}}}\limits_{x \to a} g\left( x \right)\) exists but \(g\left( a \right)\) is not defined.

(c) \(\mathop {{\bf{lim}}}\limits_{x \to {a^ - }} g\left( x \right)\) and \(\mathop {{\bf{lim}}}\limits_{x \to {a^ + }} g\left( x \right)\) both exists but \(\mathop {{\bf{lim}}}\limits_{x \to a} g\left( x \right)\) does not exist.

(d) \(\mathop {{\bf{lim}}}\limits_{x \to {a^ + }} g\left( x \right) = g\left( a \right)\) but \(\mathop {{\bf{lim}}}\limits_{x \to {a^ - }} g\left( x \right) \ne g\left( a \right)\).

Calculate each of the limits

limh→04(1+h)2-4h

19-32: Prove the statement using the \(\varepsilon ,\delta \) definition of a limit.

31. \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 1} \right) = 3\)

The point \(P\left( {{\bf{1}},{\bf{0}}} \right)\) lies on the curve \(y = {\bf{sin}}\left( {\frac{{{\bf{10}}\pi }}{x}} \right)\).

a. If Qis the point \(\left( {x,{\bf{sin}}\left( {\frac{{{\bf{10}}\pi }}{x}} \right)} \right)\), find the slope of the secant line PQ (correct to four decimal places) for \(x = {\bf{2}}\), 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes appear to be approaching a limit?

b. Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the slope of the tangent line at P.

c. By choosing appropriate secant lines, estimate the slope of the tangent line at P.
See all solutions

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