Minimizing the slope of the tangent to a curve

In summary: So the critical value is $x=\frac{8}{6}=\frac{4}{3}$.In summary, the point on the graph of y = x3 - 4x2 where the tangent line has the minimum slope is (4/3, -128/27). To find this point, the slope of the curve was first derived, resulting in 3x2-8x. Setting this equal to 0 and solving for x, the critical value was found to be x = 4/3. The corresponding y-value was then found by plugging in x = 4/3 into the original function, resulting in y = -128/27.
  • #1
leprofece
241
0
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
 
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  • #2
Re: tangente to a curve

leprofece said:
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 $0$?
 
  • #3
Re: tangente to a curve

caffeinemachine said:
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
 
  • #4
Re: tangente to a curve

leprofece said:
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
 
  • #5
Re: tangente to a curve

MarkFL said:
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
 
  • #6
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?
 
  • #7
MarkFL said:
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
 
  • #8
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:

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

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?
 
  • #9
MarkFL said:
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:

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

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
 
  • #10
leprofece said:
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 $6x-8=0$.
 

FAQ: Minimizing the slope of the tangent to a curve

How do you minimize the slope of a tangent to a curve?

To minimize the slope of a tangent to a curve, you need to find the point where the derivative of the curve is equal to zero. This point is called the critical point and represents the minimum or maximum value of the function at that point.

Why is minimizing the slope of a tangent important?

Minimizing the slope of a tangent is important because it allows us to find the minimum or maximum values of a function. This is useful in many fields of science, such as economics, physics, and engineering, where finding the optimal solution is crucial.

What is the relationship between the slope of a tangent and the derivative of a curve?

The slope of a tangent to a curve at a specific point is equal to the derivative of the curve at that point. This means that by finding the derivative of a curve, we can determine the slope of the tangent at any point on the curve.

Can the slope of a tangent be negative?

Yes, the slope of a tangent can be negative. This indicates that the curve is decreasing at that point. The slope of a tangent can be positive, negative, or zero, depending on whether the curve is increasing, decreasing, or neither at that point.

How does the slope of a tangent change as we move along a curve?

The slope of a tangent changes as we move along a curve because it is dependent on the shape of the curve at that point. If the curve is steep, the slope of the tangent will be large, and if the curve is flat, the slope of the tangent will be small. The slope of the tangent also changes from positive to negative or vice versa as the curve changes direction.

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