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Question: How do you evaluate \(f(x + 1)\) given the function \(f(x) = 3 - \dfrac{1}{2}x\)?...

How do you evaluate f(x+1)f(x + 1) given the function f(x)=3−12xf(x) = 3 - \dfrac{1}{2}x?

Explanation

Solution

To evaluate the given function we will substitute (x+1)(x + 1) everywhere instead of xx to get the value of f(x+1)f(x + 1) and then doing some simplify the result to get the required solution.

Complete step-by-step solution:
We have the given function as:
⇒f(x)=3−12x\Rightarrow f(x) = 3 - \dfrac{1}{2}x
Now we have to find f(x+1)f(x + 1)therefore on substitution the function becomes:
⇒f(x+1)=3−12(x+1)\Rightarrow f(x + 1) = 3 - \dfrac{1}{2}(x + 1)
On multiplying the terms, we get:
⇒f(x+1)=3−12×x−(12×1)\Rightarrow f(x + 1) = 3 - \dfrac{1}{2} \times x - \left( {\dfrac{1}{2} \times 1} \right)
On multiplying the terms, we get:
⇒f(x+1)=3−12x−(12)\Rightarrow f(x + 1) = 3 - \dfrac{1}{2}x - \left( {\dfrac{1}{2}} \right)
On opening the brackets, we get:
⇒f(x+1)=3−12x−12\Rightarrow f(x + 1) = 3 - \dfrac{1}{2}x - \dfrac{1}{2}
We can rearrange and write the expression as:
⇒f(x+1)=3−12−12x\Rightarrow f(x + 1) = 3 - \dfrac{1}{2} - \dfrac{1}{2}x
Now on taking the lowest common multiple on all the terms, we get:
⇒f(x+1)=62−12−12x\Rightarrow f(x + 1) = \dfrac{6}{2} - \dfrac{1}{2} - \dfrac{1}{2}x
Now on writing the first and second term in the form of a common denominator, we get:
⇒f(x+1)=6−12−12x\Rightarrow f(x + 1) = \dfrac{{6 - 1}}{2} - \dfrac{1}{2}x
On simplifying the expression, we get:
⇒f(x+1)=52−12x\Rightarrow f(x + 1) = \dfrac{5}{2} - \dfrac{1}{2}x
Now on writing the terms in the form of a common denominator, we get:
⇒f(x+1)=5−x2\Rightarrow f(x + 1) = \dfrac{{5 - x}}{2}

5−x2\dfrac{{5 - x}}{2}, which is the required evaluation for f(x+1)f(x + 1).

Note: To check whether the solution is correct we can cross check the value we get by substituting terms in f(x)f(x) and f(x+1)f(x + 1) respectively.
Now let’s consider x=1x = 1, on substituting it in the evaluated function, we get:
f(1+1)=5−x2f(1 + 1) = \dfrac{{5 - x}}{2}
We get:
f(2)=5−x2f(2) = \dfrac{{5 - x}}{2}
On substituting x=2x = 2, we get:
f(2)=5−22=32f(2) = \dfrac{{5 - 2}}{2} = \dfrac{3}{2}
Now on substituting x=2x = 2 in f(x)f(x), we get:
f(2)=3−12xf(2) = 3 - \dfrac{1}{2}x
On substituting the value of xx, we get:
f(2)=3−12×2f(2) = 3 - \dfrac{1}{2} \times 2
On taking the lowest-common multiple, we get:
f(2)=5−22=32f(2) = \dfrac{{5 - 2}}{2} = \dfrac{3}{2}, which is the same value therefore, the solution is correct.
It is to be remembered that these questions are of composite functions, the value which is written in the bracket of f(x)f(x) should be substituted for every occurrence of xx in the function, be it is in the form of logarithm, exponent, algebraic, trigonometric etc.
It is to be remembered that the value of the denominator should not be zero while doing sums of f(0)f(0).