Question

How do you find the value of $f\left( \dfrac{1}{2} \right)$, if $f(x)=3x-4$?

Answer 1

Given a function $y=f(x)$, for $x\in$ Real numbers. The coordinates of any point on the curve of the function are $\left( x,f(x) \right)$. This means we can find it by substituting the value of x in the equation of function and then calculating the value of function at that point.

Complete step by step answer:
The given function is $f(x)=3x-4$. We have to find the functional value at the point whose X-coordinate is $\dfrac{1}{2}$. To find the Y-coordinate or value of function at this point. We have to substitute the value of x in the equation of the given function.
The given equation of the function is $f(x)=3x-4$. Substituting $x=\dfrac{1}{2}$ in the equation of the function. We get
$\Rightarrow f\left( \dfrac{1}{2} \right)=3\left( \dfrac{1}{2} \right)-4$
Multiplying both sides of the above equation by 2, we get
$\Rightarrow 2\times f\left( \dfrac{1}{2} \right)=\dfrac{3}{2}\times 2-4\times 2$
Canceling out 2 as a common factor of numerator and denominator in RHS, we get

& \Rightarrow 2\times f\left( \dfrac{1}{2} \right)=3-8 \\\ & \Rightarrow 2\times f\left( \dfrac{1}{2} \right)=-5 \\\ \end{aligned}$$ Dividing both sides of the above equation by 2, we get $$\Rightarrow \dfrac{2\times f\left( \dfrac{1}{2} \right)}{2}=\dfrac{-5}{2}$$ Canceling out 2 as a common factor of numerator and denominator in LHS, we get $$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{-5}{2}$$ **Note:** For these types of questions, after we substitute the value of x in the equation. If the function is becoming undefined, in other words, we are getting an indeterminate form of the type $$\dfrac{0}{0}$$ or $$\dfrac{\infty }{\infty }$$, or any other form like these. In this case, we can not find the functional value. However, we can find the limit of the function by using different methods, and state whether the function is continuous or not at that point.