Can someone check my homework?

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

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1 Answer

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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.