Question

find the approximations of ^3
√7.8​

Answer 1

Answer:

Derivatives are used to find the linear approximation of a function. Linear approximation of a function is a line that can substitute it.

Answer 2

The approximate value of $\sqrt[3]{7.8}$ is  1.9834 when the 3 root is nothing but cube root which we can write as the power of ratio of 1 and 3.

Given that,

We have to find the approximation value of $\sqrt[3]{7.8}$

We know that,

Consider y = $x^{\frac{1}{3} }$

Let x = 8 and Δx = -0.2

y = $x^{\frac{1}{3} }$

Differentiate with respect to x

$\frac{dy}{dx}= \frac{1}{3}(x)^{\frac{1}{3}-1}= \frac{1}{3}(x)^{\frac{-2}{3}}= \frac{1}{3(x)^{\frac{2}{3} }}$

$\frac{dy}{dx}= \frac{1}{3(x)^{\frac{2}{3} }}$

Δy = $(\frac{dy}{dx})$Δx

Δy = $\frac{1}{3(x)^{\frac{2}{3} }}$Δx

Substituting the values x =8 and Δx = -0.2

Δy = $\frac{1}{3(8)^{\frac{2}{3} }}$(-0.2) =$\frac{-0.2}{3(2^3)^{\frac{2}{3}} }$ =  $\frac{-0.2}{12}$

Δy =  $\frac{-0.2}{12}$

So,

Δy = f(x + Δx) -f(x)

Δy = (x + Δx$)^{\frac{1}{3}}$ -(x$)^{\frac{1}{3}}$

$\frac{-0.2}{12}= (8-0.2)^{\frac{1}{3}}-(8)^\frac{1}{3}$

$\frac{-0.2}{12}= (7.8)^{\frac{1}{3}}-(2^3)^\frac{1}{3}$

$\frac{-0.2}{12}= (7.8)^{\frac{1}{3}}-2$

$2-\frac{0.2}{12}= (7.8)^{\frac{1}{3}}$

$\frac{24-0.2}{12}= (7.8)^{\frac{1}{3}}$

$\frac{23.8}{12}= (7.8)^{\frac{1}{3}}$

$\sqrt[3]{7.8}$ = 1.9834

Therefore, The approximate value of $\sqrt[3]{7.8}$ is  1.9834.

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