for the function $f(x)=4x^{2}$, make a table of slopes of secant lines and make a conjecture about the slope…

for the function $f(x)=4x^{2}$, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at $x = 2$.\n\nit can be conjectured that the slope of the tangent line at $x = 2$ is \n(round to the nearest integer as needed.)

for the function $f(x)=4x^{2}$, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at $x = 2$.\n\nit can be conjectured that the slope of the tangent line at $x = 2$ is \n(round to the nearest integer as needed.)

Answer

Answer:

16

Explanation:

Step1: Analyze the trend of secant - line slopes

As the left - hand endpoint of the interval gets closer and closer to (x = 2) (i.e., as the interval ([a,2]) becomes smaller with (a\to2^{-})), we observe the values of the slopes of the secant lines: (12) (for ([1,2])), (14) (for ([1.5,2])), (15.6) (for ([1.9,2])), (15.96) (for ([1.99,2])), (15.996) (for ([1.999,2])).

Step2: Make a conjecture

We notice that as (a) approaches (2) from the left (i.e., (a\to2^{-})), the slopes of the secant lines (m_{sec}=\frac{f(2)-f(a)}{2 - a}=\frac{4\times2^{2}-4a^{2}}{2 - a}=\frac{16 - 4a^{2}}{2 - a}=\frac{4(4 - a^{2})}{2 - a}=\frac{4(2 + a)(2 - a)}{2 - a}=4(2 + a)) (for (a\neq2)) are approaching (16).

Another way (using the limit definition of the derivative): The derivative of (y = f(x)=4x^{2}) using the power rule (y^\prime=8x). When (x = 2), (y^\prime|_{x = 2}=8\times2=16). The slope of the tangent line at (x = 2) is the limit of the slopes of the secant lines as the interval ([a,2]) shrinks to the point (x = 2).