Questions/Answers on Calculus

Running head: CALCULUS
CALCULUS
Name of the Student:
Name of the University:
Author Note:

CALCULUS1
Answer to Question 1:
Given,
The equation of the curve:
y = √(1-x) = f(x)
at the point x = 0.
Therefore, f(0) = 1
Point of the tangent is (0,1)
Now the slope of the tangent can be calculated as:
mtan = lim
h→ 0
f ( a+h ) −f ( a)
h [Using limit definition]
= lim
h→ 0
f ( 0+h ) −f (0)
h
= lim
h→ 0
√ 1−h−1
h
Now, Multiplying both numerator and denominator by (√1−h+1) we get,
mtan = lim
h→ 0
( √1−h−1)( √1−h+ 1)
h( √1−h+1)
= lim
h→ 0
( √(1−h) ) 2−12
h ( √ 1−h+1)
= lim
h→ 0
1−h−1
h ¿ ¿ ¿

CALCULUS2
= lim
h→ 0
−h
h ( √1−h+1)
= lim
h→ 0
−1
√1−h+ 1
Putting h=0 in the above equation, we get,
= −1
2
Therefore, the slope of the tangent is −1
2
Now, point is (0,1) and slope is -1/2. Therefore, the equation of the tangent is:
y – 1 = −1
2 (x - 0)
 y -1 = −x
2
 2y -2 = -x
 2y + x -2 = 0
Thus, the equation of the tangent to the curve y = √ (1-x) at x =0 is
2y + x – 2 = 0
Answer to Question 2:
Given curve is:
y = 2
3 x2−1