Applied Calculus: UNIT I: Differential Calculus
Odd and Even functions
New functions from old function | Differential Calculus
Explanation, Formula, Equation, Example and Solved Problems - Differential Calculus: New functions from old function:
ODD AND EVEN FUNCTIONS
(a) f(‒x) = (‒x)2
= x2 = f(x)
Therefore f is an even function.
(b) f(‒x) = |‒x|
= |x| = f(x)
Therefore f is an even function.
(c) f(‒x) = cos(‒x)
= cos x = f(x)
Therefore f is an even function.
(d) f(‒x) = ‒x = ‒f(x)
Therefore f is an odd function.
(e) f(‒x)=sin(‒x)
= ‒sin x = ‒ f(x)
Therefore f is an odd function.
Example 26. Determine whether each of the following functions is even,
odd, or neither even or odd.
(i) f(x) = x sinx
(ii) f(x) = xcosx
(iii) f(x) = x + x3
(iv) f(x) = 1‒x4
(v) f(x) = x5+x
(vi) f(x) = 2x ‒ x2
Solution:
(i) f(x) = ‒x sin (‒x)
= ‒x (‒ sin x) = x sinx = f(x)
Therefore f is an even function.
(ii) f(‒x) = ‒x cos (‒x)
= ‒x cos x = f(x)
Therefore f is an odd function.
(iii) f(‒x) = ‒x+(‒x)3
= ‒x‒x3 = (x + x3) = ‒f(x)
Therefore f is an odd function.
(iv) f(‒x) = 1− (−x)4
= 1− x4 = f(x)
Therefore f is an even function.
(v)ƒ(−x) = (‒x)5 ‒ x
= −x5 − x = −(x5+x) = −f(x)
Therefore f is an odd function.
(vi) f(−x) = 2(‒x) ‒ (‒x)2
= ‒2x ‒ x2 = −(2x + x2)
Since f(‒x) ≠ f(x) and f(‒x) ≠ ‒f(x), we conclude that f is neither even nor odd.
EXERCISE
8. Determine whether each of the following functions is even, odd, or neither even
or odd.
(a) f(x) = x2 + 1 Ans: f is an even function
(b) ƒ(x) = x3 + x Ans: f is an odd function
(c) ƒ (x) = 1 / x2‒1 Ans: f is an even function
(d) f(x) = 1 / x‒1 Ans: f is neither odd nor even
(e) f(x) = 2x+1 Ans: f is neither odd nor even
Applied Calculus: UNIT I: Differential Calculus : Tag: Applied Calculus : New functions from old function | Differential Calculus - Odd and Even functions
Applied Calculus: UNIT I: Differential Calculus
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