Find the linearization L(x) of the function at a. f(x) = x^4 + 2x^2, a = 1
1 answer:
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
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Answer:
- (0, 5√3/2 + 2)
Step-by-step explanation:
Given is the absolute value function.
<u>Observations:</u>
- It has a slope of ±√3 and the y- intercept of 2.
- There is no horizontal shift, so the the y-axis is the line of symmetry.
- The y-axis is also an angle bisector of the two lines.
- The foot P₁P₂ is parallel to the x-axis since it's perpendicular to the y- axis.
We need to find the coordinates of intersection of the line P₁P₂ with the y- axis (the point Y in the picture).
Consider the triangle AYP₂.
We know AP₂ = 5.
<u>The angle YAP₂ is:</u>
- arctan (1/√3) = 30°
<u>The distance AY is:</u>
- AY = AP₂ cos 30° = 5*√3/2
<u>The distance from the x-axis to the point Y is:</u>
- 5√3/2 + 2, added the y- intercept of the graphed lines
<u>The coordinates of the point Y:</u>
- (0, 5√3/2 + 2)
