Q 1 how to find maxmium nd minmium value or point in double differtiation !!!!!
Q 2 . X= 2t-t^2 + 1
Find max.& min value or points !!!!!!!!!
Answers
Here is your answer____________
Q 1. Answer : -
There are Various methods in order to find maximum or minimum value of a function.
One of the conventional methods is:
01. Find the derivative of the function and equate it to zero.
02. Find the roots of the differentiated equation.
03. Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.
04. If the value comes out to be negative, At the particular value of the root Maximum occurs.
05. Then substitute the value in original expression to get Maximum of the function.
06. If the value of double derivative after substituting the root is positive, Minimum occurs.
07. Then substitute the value in original equation to get Minimum value of the function.
08. If the Second derivative is Zero:
Then go for higher derivatives of the function & substitute the value of the root in the nth order derivative expression.
09. If it's positive it would give theMaximum of the function at the particular root.
________________________
Q 2. Answer
given,
X = 2t - t^2 + 1
let, f(x) = 2t - t^2 + 1
differentiate with respect to t
f'(x) = 2 - 2t
f'(x) = 2(1 - t)
we know that,
for the maximum or minimum
f'(x) = 0
ie. 2(1 - t) = 0
1 - t = 0
t = 1
Now,
angain differentiate f'(x) with respect to t
we get,
f"(x) = -t
f"(x) = -(1)
f"(x) = -1 < 0
hence f"(x) is less than 0
f(x) is maximum at t = 1.
And the maximum value of f(x) is,
f(1) = 2(1) - 2^1 + 1
= 2-2+1
=1
---------------------
Hence given X is maximum at t=1 and maximum value of X is 1.
_________________________
Hope this answer will help u....
Answer:
Q 1. Answer : -
There are Various methods in order to find maximum or minimum value of a function.
One of the conventional methods is:
01. Find the derivative of the function and equate it to zero.
02. Find the roots of the differentiated equation.
03. Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.
04. If the value comes out to be negative, At the particular value of the root Maximum occurs.
05. Then substitute the value in original expression to get Maximum of the function.
06. If the value of double derivative after substituting the root is positive, Minimum occurs.
07. Then substitute the value in original equation to get Minimum value of the function.
08. If the Second derivative is Zero:
Then go for higher derivatives of the function & substitute the value of the root in the nth order derivative expression.
09. If it's positive it would give theMaximum of the function at the particular root.
________________________
Q 2. Answer
given,
X = 2t - t^2 + 1
let, f(x) = 2t - t^2 + 1
differentiate with respect to t
f'(x) = 2 - 2t
f'(x) = 2(1 - t)
we know that,
for the maximum or minimum
f'(x) = 0
ie. 2(1 - t) = 0
1 - t = 0
t = 1
Now,
angain differentiate f'(x) with respect to t
we get,
f"(x) = -t
f"(x) = -(1)
f"(x) = -1 < 0
hence f"(x) is less than 0
f(x) is maximum at t = 1.
And the maximum value of f(x) is,
f(1) = 2(1) - 2^1 + 1
= 2-2+1
=1
---------------------
Hence given X is maximum at t=1 and maximum value of X is 1.