Question

Calc help pretty please

Answer 1

A

Explanation:

Linear approximation is basically using the equation of a tangent line to approximate values of f.

For example

we have function

`(x + 1) {}^(2)`

First, things first, let take the derivative to get the gradient function.

`(d)/(dx) (x + 1) {}^(2) = 2(x + 1)(1) = 2(x + 1)`

Note: I used the chain rule.

so we get

`2(x + 1)`

Now, notice the formula for tangent approximation.

`f(x) = f(a) + f'(a)(x - a)`

Here x=7.8

a is 8 so we get

`f(7.8) = f(8) + f'(8)(7.8 - 8)`

To find f(8), plug 8 into the original function for x.

`(8 + 1) {}^(2) = 81`

To find f'(8), plug 8 into the derivative function.

`2(8 + 1) = 18`

So we get

`f(7.8) = 81 + 18( - 0.2)`

`f(7.8) = 81 - 3.6`

`f(7.8) = 77.4`

Answer 2

A. 77.4

Explanation:

Linear approximation formula

`L(x)=f(a)+f'(a)(x-a)`

Given function:

`f(x)=(x+1)^2`

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}* \frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

`\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=xn^(n-1)$\\\end{minipage}}`

`\boxed{\begin{minipage}{4 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\frac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}`

Use the chain rule to differentiate the function.

`\textsf{Let }\:y=u^2\:\textsf{ where }u=(x+1)`

Differentiate the two parts separately:

`y=u^2 \implies \frac{\text{d}y}{\text{d}u}=2u`

`u=x+1 \implies \frac{\text{d}u}{\text{d}x}=1`

Put everything back into the chain rule formula:

`\begin{aligned} \implies \frac{\text{d}y}{\text{d}x} & =2u * 1\\ & = 2u \\ & = 2(x+1)\\ & = 2x+2 \end{aligned}`

`\textsf{Therefore, }\:f'(x)=2x+2.`.

The linear approximation at a = 8 is:

`\begin{aligned}L(x) & =f(a)+f'(a)(x-a)\\\\\implies L(x) & = f(8)+f'(8)(x-8)\\& = (8+1)^2+(2(8)+2)(x-8)\\& = 81+18(x-8)\\& = 18x-63\end{aligned}`

Finally, substitute x = 7.8 into the linear approximation equation:

`\begin{aligned}\implies L(7.8) & =18(7.8)-63\\& = 140.4-63\\& = 77.4\end{aligned}`