Can someone check my homework?

f(x)= (x-2)^(2) (x-5)

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1 Answer
May 20, 2018

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Explanation:

i).

We are given f(x)=(x-2)^2(x-5) and x=1 ,x=5

First find the y coordinates by plugging x into f(x)

x=1

f(1)=((1)-2)^2((1)-5)=-4

Coordinates: color(blue)((1,-4)

x=5

f(5)=((5)-2)^2((5)-5)=0

Coordinates: color(blue)((5,0)

These are the endpoints of the line segment.

Gradient is:

"change in y"/("change in x")=(y_2-y_1)/(x_2-x_1)

:.

(0-(-4))/(5-1)=4/4=1

Gradient is 1

ii)

The coordinate of the midpoint are given by:

((x_1+x_2)/2,(y_1+y_2)/2)

((1+5)/2,(0-4)/2)->color(blue)((3,-2)

If this point lies of f(x) then:

f(3)=-2

f(3)=((3)-2)^2((3)-5)=-2

This lies on f(x)

iii)

If the gradients are are equal ie 1, the simplest way to find these is to differentiate f(x):

expanding f(x):

f(x)=x^3-9x^2+24x-20

dy/dx( x^3-9x^2+24x-20)=3x^2-18x+24

Since this is a gradient function and we need the gradient to be 1:

3x^2-18x+24=1

3x^2-18x+23=0

We solve this for x:

Using quadratic formula give:

x=(9+2sqrt(3))/3, x=(9-2sqrt(3))/3

These are the values of x for which the tangent is 1:

This is confirmed by the graph:

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I don't know if you are supposed to use calculus methods to answer iii. This is the way I would solve it.

You seem to have the wrong results for i and ii. I don't know how you got f(1)=24, this would have made ii wrong as well.

Hope this helps.