Question

Suppose that we don't have a formula for g ( x ) g(x) but we know that g ( 1 ) = − 1 g(1)=-1 and g ' ( x ) = √ x 2 + 15 g′(x)=x2+15 for all x x. (a) Use a linear approximation to estimate g ( 0.9 ) g(0.9) and g ( 1.1 ) g(1.1). (Round your answers to two decimal places.) g ( 0.9 ) ≈ g(0.9)≈ g ( 1.1 ) ≈ g(1.1)≈ (b) Are your estimates in part (a) too large or too small? Explain. The slopes of the tangent lines are positive and the tangents are getting steeper, so the tangent lines lie below the curve. Thus, the estimates are too small. The slopes of the tangent lines are positive and the tangents are getting steeper, so the tangent lines lie above the curve. Thus, the estimates are too large. The slopes of the tangent lines are positive but the tangents are becoming less steep, so the tangent lines lie below the curve. Thus, the estimates are too small. The slopes of the tangent lines are positive but the tangents are becoming less steep, so the tangent lines lie above the curve. Thus, the estimates are too large.

Answer 1

Answer:

g(0.9) ≈ -2.6

g(1.1) ≈ 0.6

For 1.1 the estimation is a bit too high and for 0.9 it is too low.

Step-by-step explanation:

For values of x near 1 we can estimate g(x) with t(x) = g'(1) (x-1) + g(1). Note that g'(1) = 1²+15 = 16, and for values near one g'(x) is increasing because x² is increasing for positive values. This means that the tangent line t(x) will be above the graph of g, and the estimates we will make are a bit too big for values at the right of 1, like 1.1, and they will be too low for values at the left like 0.9.

For 0.9, we estimate

g(0.9) ≈ 16* (-0.1) -1 = -2.6

g(1.1) ≈ 16* 0.1 -1 = 0.6