estimate the instantaneous rate of change of $g(x)=3x^{2}+3$ at the point: $x = 3$. in other words, choose x…

estimate the instantaneous rate of change of $g(x)=3x^{2}+3$ at the point: $x = 3$. in other words, choose x - values that are getting closer and closer to 3 and compute the slope of the secant lines at each value. then, use the trend/pattern you see to estimate the slope of the tangent line. your answer should be accurate to at least 2 decimal places. question help: message instructor submit question jump to answer

estimate the instantaneous rate of change of $g(x)=3x^{2}+3$ at the point: $x = 3$. in other words, choose x - values that are getting closer and closer to 3 and compute the slope of the secant lines at each value. then, use the trend/pattern you see to estimate the slope of the tangent line. your answer should be accurate to at least 2 decimal places. question help: message instructor submit question jump to answer

Answer

Explanation:

Step1: Recall the slope - formula for secant line

The slope of the secant line between two points $(x_1, g(x_1))$ and $(x_2, g(x_2))$ on the curve $y = g(x)$ is $m=\frac{g(x_2)-g(x_1)}{x_2 - x_1}$. Let $x_1 = 3$ and choose $x_2$ values close to 3.

Step2: Calculate $g(x)$

Given $g(x)=3x^{2}+3$. Then $g(3)=3\times3^{2}+3=3\times9 + 3=27 + 3=30$.

Step3: Choose $x_2 = 3.01$

$g(3.01)=3\times(3.01)^{2}+3=3\times9.0601+3=27.1803 + 3=30.1803$. The slope of the secant line between $(3,g(3))$ and $(3.01,g(3.01))$ is $m_1=\frac{g(3.01)-g(3)}{3.01 - 3}=\frac{30.1803 - 30}{0.01}=\frac{0.1803}{0.01}=18.03$.

Step4: Choose $x_2 = 2.99$

$g(2.99)=3\times(2.99)^{2}+3=3\times8.9401+3=26.8203+3=29.8203$. The slope of the secant line between $(3,g(3))$ and $(2.99,g(2.99))$ is $m_2=\frac{g(3)-g(2.99)}{3 - 2.99}=\frac{30 - 29.8203}{0.01}=\frac{0.1797}{0.01}=17.97$.

Step5: Estimate the instantaneous rate of change

As $x_2$ gets closer to 3, the slopes of the secant lines are approaching 18.00.

Answer:

18.00