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:
$55.80
Step-by-step explanation:
8, 16, 24, 32, 36
8=12.40, 16=24.80, 24=37.2, 32=49.6,
+4 makes it 36 so it's have of 8, so add 6.20 to the end and it's 55.80
Answer:
(16-2)/7
Step-by-step explanation:
16-2=14
14/7=2
Equation: (16-2)/7
Answer:
(x - 6)^2 = 42
Step-by-step explanation:
Using " ^ " to indicate exponentiation, we have:
x^2 - 12x -6=0
Here the coefficients of this quadratic equation are a = 1, b = -12 and c = -6.
Start by taking HALF of the coefficient of x, that is, HALF of -12. We get -6.
Next, square this result: (-6)^2 = 36.
Finally, add this 36 to both sides, obtaining:
x^2 - 12x + 36 - 6 = 36
Rewriting the first three terms, we get:
(x - 6)^2 - 6 = 36, or
(x - 6)^2 = 42 This is the end result to "completing the square."
If you want to go further and actually solve for x, then see below:
Taking the square root of both sides, we get
x - 6 = ±√42
Then x = 6 ± √42, or
x = 6 + √42 and
x = 6 - √42