Answer:
The instantaneous rate of change of h(t) = 2t² + 2 at the point t = −1 is -4.
Step-by-step explanation:
If two distinct points [tex]P(x_1,y_1)[/tex] and [tex]Q(x_2,y_2)[/tex] lie on the curve [tex]y=f(x)[/tex], the slope of the secant line connecting the two points is
[tex]m_{sec}=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]
If we let the point [tex]x_2[/tex] approach [tex]x_1[/tex], then Q will approach P along the graph f(x). The slope of the secant line through points P and Q will gradually approach the slope of the tangent line through P as
[tex]m_{tan}= \lim_{x_2 \to x_1}\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]
And this is the instantaneous rate of change of the function f(x) at the point [tex]x_1[/tex].
From the information given, we know that the point P [tex](-1,2(-1)^2+2)=(-1,4)[/tex] lies on the curve [tex]h(t) = 2t^2 + 2[/tex].
If Q is the point [tex](t, 2t^2 + 2)[/tex] we can find the slope of the secant line PQ for the following values of t. Because we choose values that are getting closer and closer to −1.
[tex]\begin{array}{c}-0.9&-0.99&-0.999&-0.9999\\-1.1&-1.01&-1.001&-1.0001\\\end{array}\right[/tex]
Let the point P be [tex](x_2=-1, y_2=4)[/tex] and the point Q be [tex](x_1=t, y_1=2t^2+2)[/tex]. So,
[tex]m=\frac{4-(2t^2+2)}{-1-t}\\\\m=-\frac{2\left(t+1\right)\left(t-1\right)}{-1-t}\\\\m=2\left(t-1\right)[/tex]
Next, substitute the value of x in the formula of the slope
[tex]m=2(-0.9-1)=-3.8[/tex]
Do this for the other values of x.
Below, there is a table that shows the values of the slope.
From the table, as t approaches -1 from the left side (-0.9 to -0.9999), the slopes are approaching to -4 and as t approaches -1 from the right side (-1.1 to -1.0001), the slopes are approaching to -4. The value of the slope at P(-1,4) is then m = -4.
