Question

The function $f(x)=1-x^3-x^5$ is decreasing for :
a) $1 \leq x \leq 5$
b) $x \leq 1$
c) $x \geq 1$
d) All the values of x

Answer 1

Hint: 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:
 $f(x)=1-x^3-x^5$
Now we differentiate the function,
$f’(x)=\dfrac{d}{dx} f’(x) =0-3x^2-5x^4$
Now we find the values after differentiating,
$\Rightarrow f’(x) =-3x^2-5x^4$
Now we take constants from the LHS common,
$\Rightarrow f’(x) =-x^2(3+5x^2)$
f’(x) = 0 when x = 0
x = 0 is the only critical point because $3+5x^2 >0 \forall x \in R $
f is decreasing on all the x value $ x \in R $
because f’(x)<0 for all $ x \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.