Question

Suppose that we don’t have a formula for\(g\left( x \right)\)but we know\(g\left( 2 \right) = - 4\)and\(g'\left( x \right) = \sqrt {{x^2} + 5} \)for all\(x\).

(a) Use a linear approximation to estimate\(g\left( {1.95} \right)\)and\(g\left( {2.05} \right)\).

(b) Are your estimates in part (a) too large or too small? Explain.

Short answer

1. The values are \(g\left( {1.95} \right) \approx - 4.1\) and \(g\left( {2.05} \right) \approx 2.85\).
2. The slope of \(g\left( x \right)\)are positive, and the tangents to the function \(g\left( x \right)\)are becoming steeper. So the tangent lines should lie below the curve of \(g\left( x \right)\). So, the estimates in part (a) are too small.

Step-by-step solution

Find the linearization of function \(g\left( x \right)\)

It is given that \(g\left( 2 \right) = - 4\) and \(g'\left( x \right) = \sqrt {{x^2} + 5} \). Thus, \(g'\left( 2 \right) = \sqrt {{2^2} + 5} = 3\). The linearization of the function \(g\left( x \right)\)at \(x = 2\) is written as \(L\left( x \right) = g\left( 2 \right) + g'\left( 2 \right)\left( {x - 2} \right)\).On plugging the values of \(g'\left( 2 \right)\), \(g\left( 2 \right)\) in \(L\left( x \right)\) and simplifying, we obtain :

^{.\(\begin{aligned}{c}L\left( x \right) &= g\left( 2 \right) + g'\left( 2 \right)\left( {x - 2} \right)\\ &= - 4 + 3\left( {x - 2} \right)\\ &= - 4 + 3x - 6\\ &= 3x - 10\end{aligned}\)}

Find \(g\left( {1.95} \right)\) and \(g\left( {2.05} \right)\)

The value of the function\(g\left( x \right)\)at\(x = 1.95\)and \(x = 2.05\) can be approximated as\(g\left( {1.95} \right) \approx L\left( {1.95} \right)\), \(g\left( {2.05} \right) \approx L\left( {2.05} \right)\) . On plugging \(x = 1.95,\,2.05\) in \(L\left( x \right)\)and simplifying, we obtain as follows:

\(\begin{aligned}{c}g\left( {1.95} \right) \approx L\left( {1.95} \right)\\ &= 3\left( {1.95} \right) - 10\\ &= - 4.15\end{aligned}\)

\(\begin{aligned}{c}g\left( {2.05} \right) \approx L\left( {2.05} \right)\\ &= 3\left( {2.05} \right) - 10\\ &= - 3.85\end{aligned}\)

Thus, \(g\left( {1.95} \right) \approx - 4.1\) and \(g\left( {2.05} \right) \approx 2.85\).

Draw the conclusion

As \(g'\left( x \right) = \sqrt {{x^2} + 5} \), it can be concluded that \(g'\left( x \right)\) is increasing and is positive for all \(x\). This implies that the slope of \(g\left( x \right)\)are positive, and the tangents to the function \(g\left( x \right)\)are becoming steeper. So the tangent lines should lie below the curve of \(g\left( x \right)\). So, the estimates in part (a) are too small.