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PAGE: 117SET: ExercisesPROBLEM: 85
Please look in your text book for this problem Statement

(a)

The function is \small f(x)=x^{\frac{3}{2}} and solution point is \small (4,\ 8).

Graph.

Graph the function \small f(x)=x^{\frac{3}{2}}.

Plot the point \small (4,\ 8).

Observe the graph.

Consider a point on the curve such that the graph appears to be linear.

One such point is \small (3.8,\ 7.45).

Find the secant line equation.

The two points are \small (4,\ 8) and \small (3.8,\ 7.45).

Slope of the secant line is \small m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}.

\small \\m=\frac{8-7.45}{4-3.8}\\ \\=\frac{0.55}{0.2}\\ \\=2.75

Slope of the secant line is \small m=2.75.

Point-Slope form of line equation: \small (y-y_{1})=m(x-x_{1}).

Substitute \small m=2.75 and \small (x_{1},\ y_{1})=(4,\ 8) in point-slope form.

\small \\(y-8)=2.75(x-4)\\ \\(y-8)=2.75x-11\\ \\y=2.75x-11+8\\ \\y=2.75x-3

Secant line equation is \small y=2.75x-3.

(b)

Equation of the tangent line is \small T(x)=f\\\'(4)(x-4)+f(4).

Consider \small f(x)=x^{\frac{3}{2}}.

Apply derivative on each side with respect to \small x.

Substitute \small x=4 in \small f\\\'(x).

\small \\f\\\'(4)=\frac{3}{2}\left (4^\frac{1}{2} \right )\\ \\=\frac{3}{2}\left ( 2 \right )\\ \\=3

The point is \small (4,\ 8) which means that \small f(4)=8.

Tangent line equation is \small T(x)=f\\\'(4)(x-4)+f(4).

Substitute \small \\f\\\'(4)=3 and \small f(4)=8 in the tangent line equation.

\small \\T(x)=3(x-4)+8\\ \\T(x)=3x-12+8\\ \\T(x)=3x-4

Tangent line equation is \small T(x)=3x-4.

Secant line equation is \small y=2.75x-3.

The secant line and tangent line appears to be same when the two points come closer.

Hence, the slope of the secant line approaches to tangent line at \small (4,\ 8), as points come closer to \small (4,\ 8).

(c)

Graph the function \small f(x)=x^{\frac{3}{2}}.

Graph the tangent line \small T(x)=3x-4.

The tangent line \small T is the most accurate tangency point.

If the point of the tangency is moved, the approximation will become less accurate.

(d)

Complete the table.

The function is \small f(x)=x^{\frac{3}{2}}.

The tangent line equation is \small T(x)=3x-4.

\small \Delta x \small -3 \small -2 \small -1 \small -0.5 \small -0.1 \small 0 \small 0.1 \small 0.5 \small 1 \small 2 \small 3
\small f(\Delta x+4) \small 1 \small 2.8284 \small 5.19615 \small 6.5479 \small 7.701 \small 8 \small 8.3018 \small 9.5459 \small 11.1803 \small 14.8969 \small 18.520
\small T(\Delta x+4) \small -1 \small 2 \small 5 \small 6.5 \small 7.7 \small 9 \small 8.3 \small 9.5 \small 11 \small 14 \small 17

Observe the table.

We can conclude that as the point moves away, the accuracy of the approximation becomes less.

(a)

The graph is

Approximating point is \small (3.8,\ 7.45).

Secant line equation is \small y=2.75x-3.

(b)

The slope of the secant line approaches to tangent line at \small (4,\ 8), as points come closer to \small (4,\ 8).

(c)

Graph of the function and tangent line is

If the point of the tangency is moved, the approximation will become less accurate.

(d)

\small \Delta x \small -3 \small -2 \small -1 \small -0.5 \small -0.1 \small 0 \small 0.1 \small 0.5 \small 1 \small 2 \small 3
\small f(\Delta x+4) \small 1 \small 2.8284 \small 5.19615 \small 6.5479 \small 7.701 \small 8 \small 8.3018 \small 9.5459 \small 11.1803 \small 14.8969 \small 18.520
\small T(\Delta x+4) \small -1 \small 2 \small 5 \small 6.5 \small 7.7 \small 9 \small 8.3 \small 9.5 \small 11 \small 14 \small 17

As the point moves away, the accuracy of the approximation becomes less.



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