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Find an equation of the tangent line to the graph of \(y = B\left( x \right)\)at\(x = 6\),if\(B\left( {\bf{6}} \right) = {\bf{0}}\),and \(B'\left( 6 \right) = - \frac{1}{2}\).

Short Answer

Expert verified

The equation of the tangent line is \(y = - \frac{1}{2}x + 3\).

Step by step solution

01

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

02

The value of the function at a given point

Suppose\(y = f\left( x \right)\) is a function. So, \(f\left( a \right) = b\) implies that the graph of the function passes through the point \(\left( {a,b} \right)\).

Now, it is given that \(B\left( 6 \right) = 0\). So, the graph of \(y = B\left( x \right)\) passes through the point \(\left( {6,0} \right)\).

03

Slope of the tangent line

The value of the derivative of a function at some given point is equal to the value of the slope of the tangent line at that point.

Now, it is given that \(B'\left( 6 \right) = - \frac{1}{2}\). So, the slope of the tangent line at \(x = 6\) is equal to \( - \frac{1}{2}\).

04

Equation of the tangent line at \(x = 6\)

Substitute the value \(\left( {{x_1},{y_1}} \right) = \left( {6,0} \right)\) and \(m = - \frac{1}{2}\) in the equation of the line and solve as follows:

\(\begin{aligned}y - {y_1} &= m\left( {x - {x_1}} \right)\\y - 0 &= - \frac{1}{2}\left( {x - 6} \right)\\y &= - \frac{1}{2}x + 3\end{aligned}\)

Thus, the equation of the tangent line is \(y = - \frac{1}{2}x + 3\).

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Most popular questions from this chapter

Sketch the graph of the function f for which \(f\left( {\bf{0}} \right) = {\bf{0}}\), \(f'\left( {\bf{0}} \right) = {\bf{3}}\), \(f'\left( {\bf{1}} \right) = {\bf{0}}\), and \(f'\left( {\bf{2}} \right) = - {\bf{1}}\).

Use thegiven graph of f to state the value of each quantity, if it exists. If it does not exists, explain why.

(a)\(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{2}}^ - }} f\left( x \right)\)

(b)\(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{2}}^ + }} f\left( x \right)\)

(c) \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{2}}} f\left( x \right)\)

(d) \(f\left( {\bf{2}} \right)\)

(e) \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{4}}} f\left( x \right)\)

(f) \(f\left( {\bf{4}} \right)\)

43-45 Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f?

44. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{{\bf{2}}^x}}&{{\bf{if}}\,\,\,x \le {\bf{1}}}\\{{\bf{3}} - x}&{{\bf{if}}\,\,\,{\bf{1}} < x \le {\bf{4}}}\\{\sqrt x }&{{\bf{if}}\,\,\,x > {\bf{4}}}\end{array}} \right.\)

41-42 Show that f is continuous on \(\left( { - \infty ,\infty } \right)\).

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{\bf{sin}}\,x}&{{\bf{if}}\,\,x < \frac{\pi }{{\bf{4}}}}\\{{\bf{cos}}\,x}&{{\bf{if}}\,\,\,x \ge \frac{\pi }{{\bf{4}}}}\end{array}} \right.\)

Each limit represents the derivative of some function f at some number a. State such as an f and a in each case.

\(\mathop {{\bf{lim}}}\limits_{x \to \frac{{\bf{1}}}{{\bf{4}}}} \frac{{\frac{{\bf{1}}}{x} - {\bf{4}}}}{{x - \frac{{\bf{1}}}{{\bf{4}}}}}\)

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