Equation of the tangent line at the indicated point

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the function y = f(x) = x^(3/4) at the point (6, 54). Participants are examining the derivative and its implications for the tangent line's equation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivative of the function, with some questioning the correctness of the derivative calculation and its evaluation at x = 6. There is confusion regarding the interpretation of the function's notation, leading to differing derivative results.

Discussion Status

There is an active exploration of the derivative's correctness and its application to find the tangent line's equation. Some participants have provided guidance on the relationship between the slope and the point on the line, while others are clarifying the function's notation.

Contextual Notes

Participants note ambiguity in the function's notation, which affects the derivative calculation. This has led to varied interpretations and results in the discussion.

carlarae
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Homework Statement


Find an equation of the tangent line at the indicated point on the graph of the function.
y=f(x)=x^3/4 , (x,y)=(6,54)


Homework Equations





The Attempt at a Solution



I did the derivative which I get 3x^2/4 and then I plugged in the 6 and get 162. Is that the whole answer? right answer? it's asking for an equation and 162 doesn't look like an equation to me.
 
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carlarae said:

Homework Statement


Find an equation of the tangent line at the indicated point on the graph of the function.
y=f(x)=x^3/4 , (x,y)=(6,54)


Homework Equations





The Attempt at a Solution



I did the derivative which I get 3x^2/4 and then I plugged in the 6 and get 162.
This is wrong. Show us how you got that number.
carlarae said:
Is that the whole answer? right answer?
No and no. The question asks for the equation of the tangent line to the curve at the point (6, 54). If you know the slope m of a line and a point (x0, y0) on it, the equation of the line is y - y0 = m(x - x0).
carlarae said:
it's asking for an equation and 162 doesn't look like an equation to me.
 
So is the derivative of x^3/4 not 3x^2/4? That will make a big difference for me to take another crack at this.
 
Welcome to PF, carlarae! :smile:

carlarae said:
So is the derivative of x^3/4 not 3x^2/4? That will make a big difference for me to take another crack at this.

Hmm, if I fill in x=6 in 3x^2/4 I get a different result...

But yes, the derivative of [itex]x^3 \over 4[/itex] is [itex]3x^2 \over 4[/itex].
 
carlarae said:
So is the derivative of x^3/4 not 3x^2/4? That will make a big difference for me to take another crack at this.
Sorry I wasn't more specific. As I like Serena points out, your derivative is fine, but the value you got isn't.
 
Is the function
[tex]\frac{x^3}{4}[/tex]
or
[tex]x^{\frac{3}{4}}[/tex]?
what you wrote was ambiguous. If the function is the first, then the derivative is
[tex]\frac{3}{4}x^2[/tex]
if the second, then the derivative is
[tex]\frac{3}{4}x^{-1/4}= \frac{3}{4x^{1/4}}[/tex]
 
HallsofIvy said:
Is the function
[tex]\frac{x^3}{4}[/tex]
or
[tex]x^{\frac{3}{4}}[/tex]?
what you wrote was ambiguous.
It's the first. What she wrote actually isn't ambiguous, if you allow for exponentiation being higher in precedence than multiplication or division.
 

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