Just a clarification: tangent line w/ smallest slope on....

[fred2028]

The question reads:

Determine the equation of the line tangent to f (x) 4x^3 + 12x^2 -96x with the smallest slope on the interval 4 ? x ? 2.

I'm just wondering, to find that, I need the inflection point where it changes from concave down to concave up right? Because when it's CD, the slopes are more and more negative, until it reaches that inflection point, where the slope begins to increase.

 

[galactus]

Re: Just a clarification

What are those little 'L's' all about?. I can't tell what your expression is.

 

[fred2028]

Re: Just a clarification

What are those little 'L's' all about?. I can't tell what your expression is.

Sorry, I copied it directly from Acrobat Reader and didn't realize it messed up the formatting.

 

[stapel]

Determine the equation of the line tangent to f (x) 4x^3 + 12x^2 -96x with the smallest slope on the interval 4 ? x ? 2.

Is the function meant to be "f(x) = 4x[sup:1d4uikuk]3[/sup:1d4uikuk] + 12x[sup:1d4uikuk]2[/sup:1d4uikuk] - 96x"? (At the least, an "equals" sign is missing in the current formatting, and the spacing is odd.) Is the interval meant to be 2 < x < 4, or something else? (Since 2 is not greater than 4, the current interval obviously cannot be correct.)

to find that, I need the inflection point where it changes from concave down to concave up right?

Why? The exercise asks for the smallest slope, not for where the slope's trend changes. Take the derivative (which, if the function is a cubic, will be a quadratic), and find where the slope is least (in absolute value) on whatever the interval is meant to be.

Eliz.

 

[Dr. Flim-Flam]

f(x) = 4x^3-12x^2-96x, f ' (x) = 12x^2-24x-96, f " (x) = 24x-24, x =1
f(1) = -104, f ' (1) = -108, hence we have a point (1,-104) and slope -108.

Ergo y+104 = -108(x-1), y = -108x+4, the line with the most negative slope tangent to f(x).