15 POINTS! Please answer question 17 (C) thank you!
2 answers:
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Consider the function 
, which has derivative 
.
The linear approximation of
for some value 
within a neighborhood of 
is given by
Let
. Then 
can be estimated to be
![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since
for 
, it follows that must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function . This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
The linear approximation of
Let
Since
Indeed, the actual value is closer to the number 3.999374902...
Answer:
a) x =3 and x=5
c) x =-1 and
i) x =0 and
Step-by-step explanation:
a)
Given that the quadratic equation
x² - 8x +15 =0
The factors of 15 = 5 × 3
x² - 5x -3x +15 =0
x(x -5) -3(x-5) =0
(x-3)(x-5) =0
x =3 and x=5
c)
3 x² + 10 x +7 =0
⇒ 3x² + 3x + 7x +7 =0
⇒ 3x (x +1) +7(x+1) =0
⇒ (x+1) ( 3 x +7) =0
⇒ x =-1 and
i)
2 x ² + 9 x =0
⇒ x( 2 x + 9) =0
⇒ x =0 and 2 x = -9
⇒ x =0 and