Question

let ( f ) be a function such that at each point ( (x, y) ) on the graph of ( f ), the slope is given by ( \frac{dy}{dx}=\frac{1}{2}x-\frac{1}{4}y^{2} ). the graph of ( f ) passes through the point ( (1,-2) ) and is concave up on the interval ( 1 < x < 1.5 ). let ( k ) be the approximation for ( f(1.3) ) found by using the locally linear approximation of ( f ) at ( x = 1 ). which of the following statements about ( k ) is true?

a ( k=-2.65 ) and is an underestimate for ( f(1.3) )
b ( k=-2.65 ) and is an overestimate for ( f(1.3) )
c ( k=-2.15 ) and is an underestimate for ( f(1.3) )
d ( k=-2.15 ) and is an overestimate for ( f(1.3) )

Explanation:

Step1: Calculate the slope at \(x = 1\)

Given \(\frac{dy}{dx}=\frac{1}{2}x-\frac{1}{4}y^{2}\), when \(x = 1\) and \(y=-2\), substitute into the derivative formula:

$$
LATEXBLOCK0
$$

Step2: Use the linear approximation formula \(L(x)=y_{0}+f^{\prime}(x_{0})(x - x_{0})\)

Here \(x_{0}=1\), \(y_{0}=-2\), \(f^{\prime}(x_{0})=-\frac{1}{2}\), and \(x = 1.3\). Then

$$
LATEXBLOCK1
$$

Step3: Analyze the concavity

Since the function \(y = f(x)\) is concave - up on the interval \(1\lt x\lt1.5\), the graph of \(y = f(x)\) lies above its tangent line on this interval.

Answer:

D. \(k=-2.15\) and is an overestimate for \(f(1.3)\)