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

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

Kaylie Carroll

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
$y-{y}_{1}=m\left(x-{x}_{1}\right)$ 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

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