Question

If
$f(x) = \: \frac{1}{2} (2x {}^{3} - {x}^{2} + x)$
Then find,
$f(x + 1) - f(x)$
​

Answer 1

Given:

$\bf{f(x)=\dfrac{1}{2}(2x^{3}-x^{2}+x)}$

To Find:

• Value of f(x+1)-f(x)

​Solution:

We know that,

• (a+b)³ =a³+b³+3ab(a+b)
• (a+b)² =a²+b²+2ab

On substituting x with x+1 in the expression of f(x), we get

$\sf{f(x+1)=\dfrac{1}{2}(2(x+1)^{3}-(x+1)^{2}+x+1)}$

Now,

$\sf{f(x+1)-f(x)=\dfrac{1}{2}(2(x+1)^{3}-(x+1)^{2}+x+1)-\dfrac{1}{2}(2x^{3}-x^{2}+x)}$

$\sf{\implies\dfrac{1}{2}(2(x^{3} +1+3x(x+1))-(x^{2} +1+2x)+x+1)-\dfrac{1}{2}(2x^{3}-x^{2}+x)}$

$\sf{\implies\dfrac{1}{2}(2(x^{3} +1+3x^{2}+3x)-x^{2}-1-2x+x+1)-\dfrac{1}{2}(2x^{3}-x^{2}+x)}$

$\sf{\implies\dfrac{1}{2}(2x^{3} +2+6x^{2}+6x-x^{2}-1-2x+x+1)-\dfrac{1}{2}(2x^{3}-x^{2}+x)}$

$\sf{\implies\dfrac{1}{2}(2x^{3} +2+6x^{2}+6x-x^{2}-1-2x+x+1-2x^{3}+x^{2}-x)}$

$\sf{\implies\dfrac{1}{2}(6x^{2}+4x+2)}$

$\sf\pink{\implies3x^{2}+2x+1}$

Hence, the value of f(x+1)-f(x) is 3x²+2x+1.