How to find which of two functions is greater (or smaller)
Here is a problem that many students find difficulty in solving that if we are provided with two functions say one is F(X) and other is G(X) and we have to find whether F(X) >= G(X) or vice versa. the problem involves a good use of functions,Derivatives and applications of derivatives.
firstly we will subtract one function from other function for example
F(X) – G(X) or vice versa
assume this as a new function H(X)
H(X) = F(X) – G(X)
now we will use the concept of increasing and decreasing function to prove whether the newly formed function is increasing or decreasing.
Now comes the real twist: if the function H(X) is increasing in the domain mentioned in the question here x>=0 the the function F(X) is greater than function G(x) and vice versa. I hope now the basic concept behind this type of questions is clear.
Show that 1 + Log(x + sqrt(x^2 + 1)) >= sqrt(1 + x^2) for all x>=0.
H(X) = 1 + Log(x + sqrt(x^2 + 1)) – sqrt(1 + x^2) x>=0
F(X) = 1 + Log(x + sqrt(x^2 + 1))
G(X) = sqrt(1 + x^2)
now to find whether the function H(X) is increasing or decrasing in given domain we will differentiate the H(X) w.r.t x
H'(x) = x.(1 + (x/sqrt( x^2 + 1))/(x + sqrt(x^2 + 1) ) + Log(x + sqrt( x^2 +1)) – x/sqrt(x^2 +1)
= x/sqrt(x^2 + 1) + Log(x + sqrt( x^2 +1)) – x/sqrt(x^2 +1)
H'(X) = Log(x + sqrt(x^2 + 1)
H'(X) >= 0 (BECAUSE, Log(x + sqrt(x^2 + 1) >=0 for x>=0 given)
therefor for any x greater that zero
H(X) >= H(0) (Property of increasing function)
1 + Log(x + sqrt(x^2 + 1)) – sqrt(1 + x^2) >= 1 + 0 -1
1 + Log(x + sqrt(x^2 + 1)) >= sqrt(1 + x^2) for all x>=0
this is what we have to prove that function F(X) os greater than G(X) or not
Hope you may have understood that how to solve this kind of problems
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|I love physics and Mathematics and love to teach both subjects to high school students. I am currently perusing my Bachelor's of Engineering degree in Sydney,Nova Scotia. I am international student from India with good English communication skills and I like to help students in their problems they ...|