Dangelo Le

2022-02-12

What is the equation of the line tangent to $f\left(x\right)=2{x}^{3}-{x}^{2}+3x$ at x=3?

Branden Nelson

Beginner2022-02-13Added 12 answers

Explanation:

We calculate the first derivative

$f\left(x\right)=2{x}^{3}-{x}^{2}+3x$

${f}^{\prime}\left(x\right)=6{x}^{2}-2x+3$

The equation of the tangent is

y-f(3)=f'(3)(x-3)

$f\left(3\right)=2\cdot {3}^{3}-{3}^{2}+3\cdot 3=54-9+9=54$

${f}^{\prime}\left(3\right)=6\cdot {3}^{2}-2\cdot 3+3=54-6+3=51$

The tangent is

y-54=51(x-3)

y=51x-99

We calculate the first derivative

The equation of the tangent is

y-f(3)=f'(3)(x-3)

The tangent is

y-54=51(x-3)

y=51x-99

Kaylie Carroll

Beginner2022-02-14Added 12 answers

When finding the tangent/normal to a line, you always have to find the gradient first.

The equation:$f\left(x\right)=2{x}^{3}-{x}^{2}+3x$ when differentiated will become

${f}^{\prime}\left(x\right)=6{x}^{2}-2x+3.$

At x=3, the gradient of the curve and the tangent will be 51.

To find the equation of the tangent, you need to use the format

$y-{y}_{1}=m(x-{x}_{1})$ with m being your gradient and $x}_{1}{y}_{1$ being your x and y values. We have the x value already, so we need to find the y value by substituting the x value into the original equation.

$2\cdot {3}^{3}-{3}^{2}+3\cdot 3=54$

At (3 , 54) the equation of the tangent would be:

y-54=51(x-3)

Simplify it by expanding the brackets on the right and the final answer will be

y=51x-99

The equation:

At x=3, the gradient of the curve and the tangent will be 51.

To find the equation of the tangent, you need to use the format

At (3 , 54) the equation of the tangent would be:

y-54=51(x-3)

Simplify it by expanding the brackets on the right and the final answer will be

y=51x-99

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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