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

Hint: 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 .