Question

The range in which $y = - {x^2} + 6x - 3$ is increasing is
1)$$$$x < 3
2)$$$$x > 3
3)$$$$7 < x < 8
4)$$$$5 < x < 6

Answer 1

We have to find the range of the function $y$ for which the function is increasing . We solve this question using the concept of applications of derivatives . We first derivative $y$ with respect to x and then computing the derivative of $y$ to 0 we find the values for $x$ . Then putting the values we compute the range of the function for which it has an increasing value.

Complete step-by-step solution:
Given : $y = - {x^2} + 6x - 3$
Now we have to derivative of $y$ with respect to
Differentiating $y$ using the given rules of derivatives :
( derivative of ${x^n} = n \times {x^{(n - 1)}}$)
( derivative of constant$= 0$)
On differentiating , we get
$\dfrac{{dy}}{{dx}} = - 2x + 6$
For increasing or decreasing value of the function put $\dfrac{{dy}}{{dx}} = 0$
Putting$\dfrac{{dy}}{{dx}} = 0$, we get
$- 2x + 6 = 0$
From , this equation , we get the value of $x$
So ,
$x = 3$
Now , the interval for increasing value the first derivative of the function should be positive
So ,
$\dfrac{{dy}}{{dx}}{\text{ > }}0$
From the value of $x$ we get two intervals I.e. $\left( { - \infty ,3} \right)$ and $\left( {3,\infty } \right)$
Now , Putting one value from each interval we can get that the function is increasing for which interval
Putting $x = 0$in $\dfrac{{dy}}{{dx}}$, we get
$\dfrac{{dy}}{{dx}} = {\text{ 6}}$
$\dfrac{{dy}}{{dx}}{\text{ > }}0$
Thus for the interval $\left( { - \infty ,3} \right)$ the function is increasing .
Putting $x = 4$in$\dfrac{{dy}}{{dx}}$, we get
$\dfrac{{dy}}{{dx}} = - 2$
$\dfrac{{dy}}{{dx}} < 0$
Thus the function is decreasing for the interval $\left( {3,\infty } \right)$.
Hence the function is increasing for $x < 3$.
Thus , the correct option is $\left( 1 \right)$.

Note: Using the property of increasing and decreasing function function we can compute that for what value of $x$ the function is decreasing and for what value of $x$ the function is increasing . If the first derivative of a function is positive for a value of $x$ then the particular value of $x$ gives the minimum value of the function and vice versa .