MHB Minimizing the slope of the tangent to a curve

[leprofece]

determine the point on the graph of: y = x³ - 4 x² in which the tangent line has the minimum slope.

answer (4/3, -128/27)

ok my original idea was yo derive the curve first
3x²-8x
But when I equal to 0 I get x= 3/8

The curve would be the main
and the constrain y = mx
I tried and i couldnot find the answer

 

[caffeinemachine]

Re: tangente to a curve

determine the point on the graph of: y = x³ - 4 x² in which the tangent line has the minimum slope.

answer (4/3, -128/27)

ok my original idea was yo derive the curve first
3x²-8x
But when I equal to 0 I get x= 3/8

The curve would be the main
and the constrain y = mx
I tried and i couldnot find the answer

Why have you equated $3x^2-8x$ to $0$?

 

[leprofece]

Re: tangente to a curve

Why have you equated $3x^2-8x$ to $0$?

I derived curve first
so i got that and I equated to 0 because the derivative equated to 0 = m = tangent

 

[MarkFL]

Re: tangente to a curve

I derived curve first
so i got that and I equated to 0 because the derivative equated to 0 = m = tangent

But you want to minimize the slope of the curve, not the curve itself. :D

 

[leprofece]

Re: tangente to a curve

But you want to minimize the slope of the curve, not the curve itself. :D

Ok friend so it must be m= y/(x-x1) or maybe y/x

 

[MarkFL]

What you want to do is find an expression for the slope of the curve (the first derivative) and then minimize that...so what should you actually do here?

 

[leprofece]

What you want to do is find an expression for the slope of the curve (the first derivative) and then minimize that...so what should you actually do here?

Excuse me I am not able to now

 

[MarkFL]

You want to minimize the slope of the curve. The slope $s$ of the curve is found by differentiating with respect to the independent variable:

$$s(x)=\frac{d}{dx}\left(x^3-4x^2 \right)=3x^2-8x$$

This is what we want to minimize. So you want to differentiate this and equate the result to zero to determine the critical value. What do you find?

 

[leprofece]

You want to minimize the slope of the curve. The slope $s$ of the curve is found by differentiating with respect to the independent variable:

$$s(x)=\frac{d}{dx}\left(x^3-4x^2 \right)=3x^2-8x$$

This is what we want to minimize. So you want to differentiate this and equate the result to zero to determine the critical value. What do you find?

I DID THAT
this is the slope for x = 3/8
and then ?
I was stuck there

 

[I like Serena]

I DID THAT
this is the slope for x = 3/8
and then ?
I was stuck there

First off, the solution for $3x^2-8x=0$ is $x=0 \vee x=8/3$ instead of $x=3/8$.

More importantly, you did not take the derivative of $3x^2-8x$.
Its derivative (the 2nd derivative of the original function), is $6x-8$.
Set it equal to zero, and you'll get $6x-8=0$.