Question

if the rate of decrease of x^2/2 - 2x+5 is twice the decrease of x then​

Answer 1

Given that,

The Function is

$y=\dfrac{x^2}{2}-2x+5$

If the rate of decrease of function is twice.

$\dfrac{dy}{dt}=2(\dfrac{dx}{dt})$....(I)

We need to calculate the value of $\dfrac{dy}{dt}$

Using given function

$\dfrac{dy}{dt}=\dfrac{d(\dfrac{x^2}{2}-2x+5)}{dt}$

$\dfrac{dy}{dt}=(x-2)\dfrac{dx}{dt}$

From equation (I)

$2(\dfrac{dx}{dt})=(x-2)\dfrac{dx}{dt}$

$x=4$

Hence, The value of x is 4.