Question

Find the value of $p\left( x \right)=4{{x}^{2}}-3x+7$ at $x=1$ .
A. 6
B. 7
C. 8
D. 9

Answer 1

Hint: We have to calculate the function values of a given function at $x=1$ . We solve the given question by simply substituting the given value of $x$ in the given function and then evaluate the expression to find the final answer.

Complete step by step solution:
We are given a function and need to calculate the value of the function for the value of $x=1$ . We will be solving the given question by simply substituting the value of $x$ in the given function.
The function is $p\left( x \right)=4{{x}^{2}}-3x+7$
According to the question,
The function has to be calculated for the following domain value,
$\Rightarrow x=1$
The domain of a given function or any function, in general, can be defined as the set of those values for which the function is defined and has some real values.
Now, we need to substitute the given domain value and simplify the expression to get the answer.
For the value of $x=1$ ,
The given function is,
$\Rightarrow p\left( x \right)=4{{x}^{2}}-3x+7$
 $p\left( 1 \right)$ is the value of the given function $p\left( x \right)$ at the value of $x=1$
Substituting the given value of $x$ on both sides of the equation, we get,
$\Rightarrow p\left( 1 \right)=4{{\left( 1 \right)}^{2}}-3\left( 1 \right)+7$
Simplifying the above equation, we get,
$\Rightarrow p\left( 1 \right)=4\left( 1 \right)-3+7$
Let us evaluate the equation further.
$\Rightarrow p\left( 1 \right)=4-3+7$
$\Rightarrow p\left( 1 \right)=\left( 4-3 \right)+7$
Subtracting the terms in the brackets on the right-hand side of the equation, we get,
$\Rightarrow p\left( 1 \right)=1+7$
Adding the terms on the right-hand side, we get,
$\therefore p\left( 1 \right)=8$
$\therefore$ Option C holds correct for the given question.

So, the correct answer is “Option C”.

Note:We must remember the following points while finding the domain of any function,
1. The denominator of a fraction cannot be zero.
2. The expression under a root sign should be positive.
The range of a function can be defined as the set of all the possible outcomes of the functions for its domain. We can calculate the values of the range by substituting the values of the domain in the given function.