Question

let ( f(x)=3 - x^{2} ) and let ( a = - 1 ). complete parts (a) through (d) below.
a. find the linear approximation ( l ) for the function ( f ) at the point ( a ).
( l(x)=2x + 4 )
b. graph ( f ) and ( l ) on the same set of axes. use the graphing tool to graph the functions.
c. based on the graphs in part (b), state whether linear approximations to ( f ) near ( a ) are underestimates or overestimates.
linear approximations to ( f(x)=3 - x^{2} ) near ( a = - 1 ) are overestimates because the graph of ( l ) lies above the graph of ( f ) near ( a ).
d. compute ( f^{prime prime}(a) ) to confirm the conclusion in part (c).
the value of ( f^{prime prime}(x) ) at ( a = - 1 ) is ( f^{prime prime}(-1)=square ), which is ( square ) 0. this means that ( f ) is concave ( square ) confirming the conclusion in part (c)

Explanation:

Step1: Differentiate the function \( f(x) \)

Given \( f(x)=3 - x^{2} \), the first derivative \( f^{\prime}(x)=-2x \), and the second derivative \( f^{\prime\prime}(x)=-2 \).

Step2: Evaluate \( f^{\prime\prime}(x) \) at \( a=-1 \)

Substitute \( x = - 1 \) into \( f^{\prime\prime}(x) \). Since \( f^{\prime\prime}(x)=-2 \) for all \( x \), then \( f^{\prime\prime}(-1)=-2 \).

Answer:

The value of \( f^{\prime\prime}(x) \) at \( a = - 1 \) is \( f^{\prime\prime}(-1)=-2 \), which is \( < \) \( 0 \). This means that \( f \) is concave \( down \) confirming the conclusion in part (c).