
How do you determine the range of a quadratic function?
Answer
463.2k+ views
Hint: In this question, we have to find the range of a quadratic function. As we know, a quadratic function is a function which has one variable and has a power of 2. Also, range consists of the values, where the output of a function lies. Thus, in this problem, we let the general form of the quadratic equation , and then we will find the first derivative of the given equation. After that, we will put the derivative equal to 0, to get the value of x. Then, we will find the second derivative of the function. In the last, we will substitute the value of x in the quadratic equation, to get the required range of the solution.
Complete step-by-step solution:
According to the question, we have to find the range of a quadratic polynomial. As we know, the range is the interval where the value of output lies.
Thus, we will let the general form of the quadratic equation be , where a, b, and c are the real numbers and x is the variable. ---- (1)
Now, we will find the first derivative of equation (1) with respect to x, we get
--------- (2)
Now, we will put the value of first derivative equal to 0 that is let f’(x) = 0 to get the value of x, we get
Now, subtract b on both side in the above equation, we get
As we know, the same terms with opposite signs cancel out each other, thus we get
Now, we will divide 2a on both sides in the above equation, we get
On further simplification, we get
-------- (3)
Now, we will again find the derivative of equation (2), we get
Since, 2a is a positive number; it implies the value of the function at is minimum.
So, we will substitute the value of x from equation (3) in equation (1), we get
On solving the brackets of the above equation, we get
On further simplifying the above equation, we get
On taking the LCM of the denominator on the right-hand side in the above equation, we get
On further solving, we get
Therefore, the minimum limit of the range of the function is and maximum limit is
Thus, the range of a quadratic equation is , that is .
Note: While solving this problem, do mention the formula you are using to avoid confusion and mathematical error. Always remember that when the value of the second derivative is negative, it means the value of the function is maximum, otherwise it is minimum.
Complete step-by-step solution:
According to the question, we have to find the range of a quadratic polynomial. As we know, the range is the interval where the value of output lies.
Thus, we will let the general form of the quadratic equation be
Now, we will find the first derivative of equation (1) with respect to x, we get
Now, we will put the value of first derivative equal to 0 that is let f’(x) = 0 to get the value of x, we get
Now, subtract b on both side in the above equation, we get
As we know, the same terms with opposite signs cancel out each other, thus we get
Now, we will divide 2a on both sides in the above equation, we get
On further simplification, we get
Now, we will again find the derivative of equation (2), we get
Since, 2a is a positive number; it implies the value of the function at
So, we will substitute the value of x from equation (3) in equation (1), we get
On solving the brackets of the above equation, we get
On further simplifying the above equation, we get
On taking the LCM of the denominator on the right-hand side in the above equation, we get
On further solving, we get
Therefore, the minimum limit of the range of the function is
Thus, the range of a quadratic equation is
Note: While solving this problem, do mention the formula you are using to avoid confusion and mathematical error. Always remember that when the value of the second derivative is negative, it means the value of the function is maximum, otherwise it is minimum.
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