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Question: How do you find \(h\left( -1 \right)\) if it is given that \(h\left( x \right)=-{{x}^{2}}-3\)?...

How do you find h(−1)h\left( -1 \right) if it is given that h(x)=−x2−3h\left( x \right)=-{{x}^{2}}-3?

Explanation

Solution

In the above problem, we have given a function in x which is: h(x)=−x2−3h\left( x \right)=-{{x}^{2}}-3. Now, we are asked to find the value of h(−1)h\left( -1 \right) which we can calculate by putting x=−1x=-1 in the function h(x)h\left( x \right). By solving the function h(x)h\left( x \right) after substituting x=−1x=-1 will give h(−1)h\left( -1 \right).

Complete step by step answer:
The function in x which is given above is as follows:
h(x)=−x2−3h\left( x \right)=-{{x}^{2}}-3
Now, in the above problem, we have to find the value of h(−1)h\left( -1 \right) which can be achieved by putting x=−1x=-1 in the above equation. Now, the substitution of x=−1x=-1 in the above equation can be achieved by putting -1 in place of x in the above equation.
h(−1)=−(−1)2−3h\left( -1 \right)=-{{\left( -1 \right)}^{2}}-3
While solving the above equation, we will get the negative sign multiplied by the negative sign and we know that when two negative signs are in multiplication with each other then we get 1.
(−1)×(−1)=+1\left( -1 \right)\times \left( -1 \right)=+1
So, using the above relation in solving the function h(x)h\left( x \right) we get,
h(−1)=−1−3 ⇒h(−1)=−4 \begin{aligned} & h\left( -1 \right)=-1-3 \\\ & \Rightarrow h\left( -1 \right)=-4 \\\ \end{aligned}
From the above solution, we got the value of h(−1)h\left( -1 \right) as -4.

Hence, h(−1)=−4h\left( -1 \right)=-4.

Note: While solving the above solution when we substitute -1 in place of x in h(x)h\left( x \right) then when taking the square of number -1, we might mistakenly write the result of this square as -1 in place of 1. We are demonstrating the mistake that we are just talking about:
h(−1)=−(−1)2−3 ⇒h(−1)=−(−1)−3 \begin{aligned} & h\left( -1 \right)=-{{\left( -1 \right)}^{2}}-3 \\\ & \Rightarrow h\left( -1 \right)=-\left( -1 \right)-3 \\\ \end{aligned}
In the examination, the examiner will take advantage of such mistakes and will give you an option where you will find the result of this mistake. For e.g. if we simplify the above equation we get:
h(−1)=−(−1)−3 ⇒h(−1)=1−3=−2 \begin{aligned} & h\left( -1 \right)=-\left( -1 \right)-3 \\\ & \Rightarrow h\left( -1 \right)=1-3=-2 \\\ \end{aligned}
So, in the multiple choice question, you might see options where you see the answer as -2. Hence, beware of making such mistakes.