Question

Calc help pretty please

Answer 1

Answer:

A

Step-by-step 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.

$\frac{d}{dx} (x + 1) {}^{2} = 2(x + 1)(1) = 2(x + 1)$

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

Answer:

A.  77.4

Step-by-step explanation:

Linear approximation formula

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

Given function:

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

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

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

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

Use the chain rule to differentiate the function.

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

Differentiate the two parts separately:

  $y=u^2 \implies \dfrac{\text{d}y}{\text{d}u}=2u$

  $u=x+1 \implies \dfrac{\text{d}u}{\text{d}x}=1$

Put everything back into the chain rule formula:

$\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}x} & =2u \times 1\\ & = 2u \\ & = 2(x+1)\\ & = 2x+2 \end{aligned}$

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

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:

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