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:
(x+12)
Step-by-step explanation:
x2 + 8x - 48
x2 + 12x - 4x - 48
(x-4)(x+12)
Answer:
cos (12+10) = cos (22)
Step-by-step explanation:
Cos 12* cos 10* - sin 12* sin 10*
Appling the trig identity: cos (a + b) = cos a.cos b - sin a.sin b
inputing the values of a and b
cos (12 + 10) = cos 12cos 10 - sin 12sin 10 = cos (22)
Answer:
2x+9=12x-21
-2. -2
9=10x -21
+21 +21
30=10x
Divide by 10 on both sides
30/10
x=3
Step-by-step explanation: