Question

Use the linear approximation to find approximate values of ${\displaystyle \sqrt[4]{15}}$

Answer 1

${\displaystyle \sqrt[4]{15}={\left(15\right)}^{\frac{1}{4}}}$

f(x) = `x^(1/4), f(x₀) = ${\displaystyle {\left(16\right)}^{\frac{1}{4}}}$ = 2

We know that

f(x₀ + Δx) = f(x₀) + f’(x₀) Δx

${\displaystyle {\left(15\right)}^{\frac{1}{4}}=2+\frac{1}{4x^{\frac{1}{4}}}(-1)}$

= ${\displaystyle 2+\frac{1}{4{\left(16\right)}^{\frac{3}{4}}}(-1)}$

= ${\displaystyle 2+\frac{1}{4\times 8}(-1)}$

= ${\displaystyle 2-\frac{1}{32}}$

= 2 – 0.03125

${\displaystyle {\left(15\right)}^{\frac{1}{4}}}$ = 1.96875