Equation of the tangent line at the indicated point

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SUMMARY

The discussion focuses on finding the equation of the tangent line for the function y=f(x)=x^(3/4) at the point (6, 54). The derivative of the function is correctly identified as f'(x)=3/4*x^(2/4), but there is confusion regarding the evaluation of the derivative at x=6. The correct slope at this point is 162, but the user initially misinterprets this value as the equation of the tangent line. The correct form of the tangent line equation is y - y0 = m(x - x0), which requires both the slope and the point of tangency.

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  • Familiarity with the equation of a line in point-slope form
  • Knowledge of exponentiation and its precedence in mathematical expressions
  • Ability to evaluate functions and derivatives at specific points
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Students studying calculus, particularly those learning about derivatives and tangent lines, as well as educators looking for examples of common misunderstandings in derivative evaluation.

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 x^3 \over 4 is 3x^2 \over 4.
 
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
\frac{x^3}{4}
or
x^{\frac{3}{4}}?
what you wrote was ambiguous. If the function is the first, then the derivative is
\frac{3}{4}x^2
if the second, then the derivative is
\frac{3}{4}x^{-1/4}= \frac{3}{4x^{1/4}}
 
HallsofIvy said:
Is the function
\frac{x^3}{4}
or
x^{\frac{3}{4}}?
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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