The equation of the tangent line at x=1 can be written in point-slope form as
... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5
Answer:
(0, 1)
(7.1, 0.49)
Step-by-step explanation:
The graph of the function is attached below.
We have an exponential function:
And we have a quadratic function:
To find the solution to:
we must look in the graph for the values of x for which both curves are intercepted, these values are:
Then the solution are the following ordered pairs:
(0, 1)
(7.1, 0.49)
Answer:
Step-by-step explanation:
<h3><u>Given data:</u></h3>Theta = θ = 35°
Radius = r = 10 ft.
<h3><u>Required:</u></h3>Arc length = ?
<h3><u>Formula:</u></h3>Put the givens in the above formula