Determine the equation for the tangent line of the function that has the largest slope

prokrastinera said:
The function: f(x) = 6/x^2+3.

Had a hard time solving this. Can you solve it too with steps?
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Did you read posting guidlines? Where are you stuck? Did you try to draw the graph and a few tangent lines?
 
The first thing you would do is differentiate: \(\displaystyle f'(x)= (6x^(-2)+ 3)'= -12x^(-3)\). The tangent line to y= f(x) at \(\displaystyle x_0\) is \(\displaystyle y= (-12x_0^{-3})(x- x_0)+ 6x_0^2+ 3\). You say "has the largest slope". The slope if that gets arbitrarily large as x nears 0.

Did you mean \(\displaystyle f(x)= 6/(x^2+ 3)= 6(x^2+ 3)^{-1}\)? \(\displaystyle f'= -12x(x^2+ 3)^{-2}\). Where is that maximum?
 
HallsofIvy said:
The first thing you would do is differentiate: \(\displaystyle f'(x)= (6x^(-2)+ 3)'= -12x^(-3)\). The tangent line to y= f(x) at \(\displaystyle x_0\) is \(\displaystyle y= (-12x_0^{-3})(x- x_0)+ 6x_0^2+ 3\). You say "has the largest slope". The slope if that gets arbitrarily large as x nears 0.

Did you mean \(\displaystyle f(x)= 6/(x^2+ 3)= 6(x^2+ 3)^{-1}\)? \(\displaystyle f'= -12x(x^2+ 3)^{-2}\). Where is that maximum?
Click to expand...
Yes I mean 1590594441186.png
 
HallsofIvy said:
The first thing you would do is differentiate: \(\displaystyle f'(x)= (6x^(-2)+ 3)'= -12x^(-3)\). The tangent line to y= f(x) at \(\displaystyle x_0\) is \(\displaystyle y= (-12x_0^{-3})(x- x_0)+ 6x_0^2+ 3\). You say "has the largest slope". The slope if that gets arbitrarily large as x nears 0.

Did you mean \(\displaystyle f(x)= 6/(x^2+ 3)= 6(x^2+ 3)^{-1}\)? \(\displaystyle f'= -12x(x^2+ 3)^{-2}\). Where is that maximum?
Click to expand...
Could you write your answer on paper and post it? I really need to pass this exam... ?
 
Take the derivative. Now you have the derivative which is just a function. Now you want to know the max of this function. How do you find the max of a function???
 
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