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Margaret [11]
2 years ago
15

Calc help pretty please

Mathematics
2 answers:
Nadya [2.5K]2 years ago
7 0

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

Gre4nikov [31]2 years ago
4 0

Answer:

A.  77.4

Step-by-step explanation:

<u>Linear approximation formula</u>

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 <u>linear approximation</u> 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 <u>linear approximation equation</u>:

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

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