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Question: If \(f(x)\) is a function such that \(f''(x) + f(x) = 0\) and \[g(x) = {\left[ {f(x)} \right]^2} + {...

If f(x)f(x) is a function such that f(x)+f(x)=0f''(x) + f(x) = 0 and g(x)=[f(x)]2+[f(x)]2g(x) = {\left[ {f(x)} \right]^2} + {\left[ {f'(x)} \right]^2} and g(3)=8g(3) = 8 , then g(8)=?g(8) = ?

Explanation

Solution

In the above question we have to find the value of g(8)g(8) . we have been given the functions and their values such as f(x)+f(x)=0f''(x) + f(x) = 0 and g(x)=[f(x)]2+[f(x)]2g(x) = {\left[ {f(x)} \right]^2} + {\left[ {f'(x)} \right]^2} . So here we will first differentiate the given function which is g(x)g(x) by using the chain rule formula and then we will simplify and solve this. We will also use the basic power rule formula to solve this question.

Formula used:
ddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}
ddx(uv)=uv+vu\dfrac{d}{{dx}}(uv) = uv' + vu'

Complete answer:
Here we have f(x)+f(x)=0f''(x) + f(x) = 0 and g(x)=[f(x)]2+[f(x)]2g(x) = {\left[ {f(x)} \right]^2} + {\left[ {f'(x)} \right]^2} .
Let us differentiate the g(x)g(x) function i.e.
g(x)=[f(x)2]+[f(x)]2\Rightarrow g(x) = \left[ {f{{(x)}^2}} \right] + {\left[ {f'(x)} \right]^2} .
By comparing with the chain rule here we have
u=(f(x))2u = {\left( {f(x)} \right)^2} and
v=(f(x))2v = {\left( {f'(x)} \right)^2} .
We can see that we have power in the function. By comparing the expression with the power rule, we have n=2n = 2 .
So by applying the power rule, we can write
(f(x))2=2(f(x))21\Rightarrow {\left( {f(x)} \right)^2} = 2{\left( {f(x)} \right)^{2 - 1}}
It gives us 2f(x)2f(x) .
Similarly for (f(x))2{\left( {f'(x)} \right)^2} , we can write this value as
2(f(x))21=2f(x)\Rightarrow 2{\left( {f'(x)} \right)^{2 - 1}} = 2f'(x)
We will now calculate vv'
We should keep in mind the constant function is not considered in this part as we have already simplified the value by power rule and the value of constant is zero.
Similarly, for uu' , we can write the value:
ddx(f(x))=f(x)\Rightarrow \dfrac{d}{{dx}}(f(x)) = f'(x)
By substituting the values in the formula, we have:
g(x)=2f(x)×f(x)+2f(x)×f(x)\Rightarrow g'(x) = 2f(x) \times f'(x) + 2f'(x) \times f''(x)
We will take the common factor out and it can be written as:

Now we can see that it is given in the question that f(x)+f(x)=0f''(x) + f(x) = 0
So, we will put this value in the expression and we have:
2f(x)[0]=0\Rightarrow 2f'(x)[0] = 0
Therefore, we have the value
g(x)=0g'(x) = 0
Now we know that the derivative of any function has value zero i.e., 00 only if it is a constant function.
So, we can say that g(x)g'(x) is a constant function.
Therefore, the value will be the same also at g(8)g(8) for all values of xx .
So, we have been given the question that g(3)=8g(3) = 8 . So it will be the same value for g(8)g(8)
Hence the required answer is g(8)=8g(8) = 8

Note: We should note that a function is defined as a unique relationship between each element of one set with only one element of another set. We can represent this also as f(x)f(x) . Differentiation is an act of computing derivatives. So, when we have function y=f(x)y = f(x) ,
we can calculate the derivative of this by suing the concept of differentiation i.e.
dydx=f(x)\dfrac{{dy}}{{dx}} = f'(x)
Or, we can say
y=f(x)y' = f'(x)