Nevaeh Glass

2022-04-10

What is the equation of the normal line of f(x)=(x-1)(x+4)(x-2) at x=3?

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Beginner2022-04-11Added 14 answers

Explanation:

before differentiating , distribute the brackets.

start with$(x-1)(x+4)={x}^{2}+3x-4$

$({x}^{2}+3x-4)(x-2)={x}^{3}-2{x}^{2}+3{x}^{2}-6x-4x+8$

$f\left(x\right)={x}^{3}+{x}^{2}-10x+8$

The derivative of f(x) is the gradient of the tangent and f'(3)

the value of the tangent.

${f}^{\prime}\left(x\right)=3{x}^{2}+2x-10$

and${f}^{\prime}\left(3\right)=3{\left(3\right)}^{2}+2\left(3\right)-10=27+6-10=23=m$

For 2 perpendicular lines , the product of their gradients

is -1

let${m}_{1}=$ gradient of normal

then$m\cdot {m}_{1}=-1\to 23\cdot {m}_{1}=-1\Rightarrow {m}_{1}=-\frac{1}{23}$

$f\left(3\right)=\left(2\right)\left(7\right)\left(1\right)=14\Rightarrow (3,14)$ is normal point

equation of normal is$y-b=m(x-a)m=-\frac{1}{23},(a,b)=(3,14)$

so$y-14=-\frac{1}{23}(x-3)$

(multiply by 23 to eliminate fraction )

$23y-322=-x+3\Rightarrow 23y+x-325=0$

before differentiating , distribute the brackets.

start with

The derivative of f(x) is the gradient of the tangent and f'(3)

the value of the tangent.

and

For 2 perpendicular lines , the product of their gradients

is -1

let

then

equation of normal is

so

(multiply by 23 to eliminate fraction )

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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