the figure below shows $f(x)$ and its local linearization at $x = a$, $y = 3x - 5$. (the local linearization…

the figure below shows $f(x)$ and its local linearization at $x = a$, $y = 3x - 5$. (the local linearization is shown in blue.)\nwhat is the value of $a$?\n$a =$\nwhat is the value of $f(a)$?\n$f(a) =$\nuse the linearization to approximate the value of $f(5.3)$.\n$f(5.3) =$\nis the approximation an under- or overestimate?\n(enter under or over.)

the figure below shows $f(x)$ and its local linearization at $x = a$, $y = 3x - 5$. (the local linearization is shown in blue.)\nwhat is the value of $a$?\n$a =$\nwhat is the value of $f(a)$?\n$f(a) =$\nuse the linearization to approximate the value of $f(5.3)$.\n$f(5.3) =$\nis the approximation an under- or overestimate?\n(enter under or over.)

Answer

Explanation:

Step1: Find the value of (a)

The local linearization (y = 3x-5) and the function (f(x)) touch at (x = a). From the graph, we can see that the (x -)coordinate of the point of tangency (where the function and its linearization meet) is (a = 4). Because when (x = 4), the linear function (y=3x - 5) and the function (f(x)) intersect.

Step2: Find the value of (f(a))

Since (y = 3x-5) is the linearization of (f(x)) at (x = a), then (f(a)=3a - 5). Substitute (a = 4) into the equation (y = 3x-5). So (f(4)=3\times4-5=12 - 5=7)

Step3: Approximate (f(5.3)) using linearization

The linear approximation formula is (L(x)=f(a)+f^{\prime}(a)(x - a)). Here (a = 4), (f^{\prime}(a)) is the slope of the linearization. Since the linearization is (y = 3x-5), (f^{\prime}(4)=3). Then (L(x)=3(x - 4)+7). Substitute (x = 5.3) into (L(x)): (L(5.3)=3\times(5.3 - 4)+7=3\times1.3+7 = 3.9+7=10.9)

Step4: Determine if it is an under - or over - estimate

The function (y = f(x)) is concave up (the graph of (f(x)) lies above its tangent line for (x) near (a) in the relevant interval). So the linear approximation (tangent line approximation) is an underestimate.

Answer:

(a = 4) (f(a)=7) (f(5.3)=10.9) under