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2 years ago

Find the linearization L(x) of the function at a. f(x) = x^4 + 2x^2, a = 1

Mathematics

1 answer:

gregori [183]2 years ago

8 0

The equation of the tangent line at x=1 can be written in point-slope form as

... L(x) = f'(1)(x -1) +f(1)

The derivative is ...

... f'(x) = 4x^3 +4x

so the slope of the tangent line is f'(1) = 4+4 = 8.

The value of the function at x=1 is

... f(1) = 1^4 +2·1^2 = 3

So, your linearization is ...

... L(x) = 8(x -1) +3

... L(x) = 8x -5

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Find the equation of the line that passes through (0, -3) and is parallel to

Tresset [83]

Hey there!

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Answer:

$\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}$

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Explanation:

To find the equation of a line, we first have to determine its slope knowing that parallel lines have the same slope.

Let the line that we are trying to determine its equation be $\: \sf{d_1} \:$ and the line that is parallel to $\: \sf{d_1} \:$ be $\: \sf{d_2} \:$ .

$\sf{d_2} \:$ passes through the points (9 , 2) and (3 , -5) which means that we can find its slope using the slope formula:

$\sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{y_2} - \orange{y_1}}{\red{x_2} - \blue{x_1 }}}$

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⇒Subtitute the values :

$\sf{(\overbrace{\blue{9}}^{\blue{x_1}}\: , \: \overbrace{\orange{2}}^{\orange{y_1}}) \: \: and \: \: (\overbrace{\red{3}}^{\red{x_2}} \: , \: \overbrace{\green{-5}}^{\green{y_2}} )}$

$\implies \sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{-5} - \orange{2}}{\red{ \: \: 3} - \blue{9 }} = \dfrac{ - 7}{ - 6} = \boxed{ \bold{\dfrac{7}{6} }}}$

$\sf{\bold{The \: slope \: of \: both \: lines \: is \: \dfrac{7}{6}}}$ .

Assuming that we want to get the equation in Slope-Intercept Form, let's substitute m = 7/6:

Slope-Intercept Form:

$\sf{y = mx + b} \\ \sf{Where \: m \: is \: the \: slope \: of \: the \: line \: and \: b \: is \: the \: y-intercept.}$

$\implies \sf{y = \bold{\dfrac{7}{6}}x + b}$ $\\$

We know that the coordinates of the point (0 , -3) verify the equation since it is on the line $\: \sf{d_1} \:$ . Now, replace y with -3 and x with 0:

$\implies \sf{\overbrace{-3}^{y} = \dfrac{7}{8} \times \overbrace{0}^{x} + b} \\ \\ \implies \sf{-3 = 0 + b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \sf{\boxed{\bold{b = -3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the equation of the line $\: \bold{d_1} \:$ is $\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}$

$\\$

▪️Learn more about finding the equation of a line that is parallel to another one here:

↣brainly.com/question/27497166

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sineoko [7]

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Step-by-step explanation:

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