Question

If $ f(x) = {x^2} $ and $ h $ is not equal to zero, find $ f(1 + h) - \dfrac{{f(1)}}{h} $ . How do you solve this problem?

Answer 1

Hint: We have been given a function $ f(x) $ . Using this function we have to simplify the expression of $ f(1 + h) - \dfrac{{f(1)}}{h} $ . The independent variable in the function is replaced by another form of expression including the variable $ h $ which is a non-zero number. We can use the expression of $ f(x) $ to arrive at the required expression.

Complete step-by-step answer:
We have been given $ f(x) = {x^2} $ .
We have to find the expression of $ f(1 + h) - \dfrac{{f(1)}}{h} $ , where $ h $ is any non-zero number.
Let us denote the new function that we have to derive as $ g(h) $ .
Then, $ g(h) = f(1 + h) - \dfrac{{f(1)}}{h} $ is the function that we have to find where $ h $ is the independent variable which is not equal to zero.
We can simplify the first term $ f(1 + h) $ using the given function of $ f(x) $ . For this we replace the independent variable $ x $ with the expression $ \left( {1 + h} \right) $ .
 $
  f(x) = {x^2} \\
   \Rightarrow f\left( {1 + h} \right) = {\left( {1 + h} \right)^2} \;
  $
Now in the second term we have to find the value of $ f(1) $ . This we can again find using the given function $ f(x) = {x^2} $ where $ x = 1 $ .
 $
  f(x) = {x^2} \\
   \Rightarrow f\left( 1 \right) = {\left( 1 \right)^2} = 1 \;
  $
Thus, we get the resulting expression as,
 $
  g(h) = f\left( {1 + h} \right) - \dfrac{{f(1)}}{h} \\
   \Rightarrow g(h) = {\left( {1 + h} \right)^2} - \dfrac{1}{h} \\
   \Rightarrow g(h) = 1 + {h^2} + 2h - \dfrac{1}{h} \\
   \Rightarrow g(h) = {h^2} + 2h + 1 - \dfrac{1}{h} \;
  $
Hence, the expression of $ f(1 + h) - \dfrac{{f(1)}}{h} $ with $ f(x) = {x^2} $ results as $ \left( {{h^2} + 2h + 1 - \dfrac{1}{h}} \right) $ where $ h $ is non-zero.
So, the correct answer is “$\left( {{h^2} + 2h + 1 - \dfrac{1}{h}} \right) $”.

Note: When we use an expression to find the value of a function we simply substitute the independent variable with the expression. The resulting expression here is in terms of $ h $ and not in terms of $ x $ as the required expression takes only $ h $ as the input variable. We have been writing $ h $ is non-zero all along the solution because we have $ h $ in the denominator in the expressions and $ h = 0 $ would make the expression indeterminate.