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Question: The function \(f(x)=1-x^3-x^5\) is decreasing for : a) \(1 \leq x \leq 5\) b) \(x \leq 1\) c)...

The function f(x)=1−x3−x5f(x)=1-x^3-x^5 is decreasing for :
a) 1≤x≤51 \leq x \leq 5
b) x≤1x \leq 1
c) x≥1x \geq 1
d) All the values of x

Explanation

Solution

A function with a graph that moves downward as it is followed from left to right is known as decreasing function. Differentiate the function and then equate with 0 to find the values.

Complete step-by-step answer:
f(x)=1−x3−x5f(x)=1-x^3-x^5
Now we differentiate the function,
f’(x)=ddxf’(x)=0−3x2−5x4f’(x)=\dfrac{d}{dx} f’(x) =0-3x^2-5x^4
Now we find the values after differentiating,
⇒f’(x)=−3x2−5x4\Rightarrow f’(x) =-3x^2-5x^4
Now we take constants from the LHS common,
⇒f’(x)=−x2(3+5x2)\Rightarrow f’(x) =-x^2(3+5x^2)
f’(x) = 0 when x = 0
x = 0 is the only critical point because 3+5x2>0∀x∈R3+5x^2 >0 \forall x \in R
f is decreasing on all the x value x∈Rx \in R
because f’(x)<0 for all x∈Rx \in R
Therefore , the correct option is d) all the values of x

Note: We need to differentiate the function and not integrate it. Students often go wrong in this step. Now students need to equate the differentiated equation with 0 to find the values of x. This step is very important as otherwise students would not be able to find the values for x.