## Precalculus (6th Edition) Blitzer

The required solution is $4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$.
Let us consider the provided ratio: $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Now, put the provided function in the above ratio: \begin{align} & \frac{\left( {{\left( x+h \right)}^{4}}+7 \right)-\left( {{x}^{4}}+7 \right)}{h}=\frac{{{\left( x+h \right)}^{4}}+7-{{x}^{4}}-7}{h} \\ & =\frac{{{\left( x+h \right)}^{4}}-{{x}^{4}}}{h} \\ & =\frac{{{x}^{4}}+4{{x}^{3}}h+6{{x}^{2}}{{h}^{2}}+4x{{h}^{3}}+{{h}^{4}}-{{x}^{4}}}{h} \\ & =\frac{h\left( 4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}} \right)}{h} \end{align} So, $\frac{\left( {{\left( x+h \right)}^{4}}+7 \right)-\left( {{x}^{4}}+7 \right)}{h}=4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$ Thus, the ratio is $4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$. It is calculated by putting the function into the given ratio and the value obtained matches with the binomial expansion formula.