Question

let ( g ) be a differentiable function such that ( g(3)=2 ) and ( g^{prime}(3)=\frac{1}{2} ). the graph of ( g ) is concave down on the interval ( 2,4 ). which of the following is true about the approximation for ( g(2.6) ) found using the line tangent to the graph of ( g ) at ( x = 3 )?

a ( g(2.6) approx 1.7 ) and this approximation is an overestimate of the value of ( g(2.6) )

b ( g(2.6) approx 1.7 ) and this approximation is an underestimate of the value of ( g(2.6) )

c ( g(2.6) approx 2.3 ) and this approximation is an overestimate of the value of ( g(2.6) )

d ( g(2.6) approx 2.3 ) and this approximation is an underestimate of the value of ( g(2.6) )

Explanation:

Step1: Find the linear approximation formula

The linear approximation formula is \(L(x)=g(a)+g^{\prime}(a)(x - a)\). Here \(a = 3\), \(g(3)=2\), \(g^{\prime}(3)=\frac{1}{2}\), and \(x = 2.6\).

$$L(2.6)=g(3)+g^{\prime}(3)(2.6 - 3)$$

Step2: Substitute the values

Substitute \(g(3) = 2\) and \(g^{\prime}(3)=\frac{1}{2}\) into the formula:

$$L(2.6)=2+\frac{1}{2}(2.6 - 3)=2+\frac{1}{2}\times(- 0.4)=2-0.2 = 1.8\approx1.7$$

Step3: Analyze the concavity

Since the graph of \(y = g(x)\) is concave - down on the interval \((2,4)\), the tangent line lies above the graph of the function. So the linear approximation is an over - estimate.

Answer:

A. \(g(2.6)\approx1.7\) and this approximation is an overestimate of the value of \(g(2.6)\)