Solving for the Tangent Line at (1,e^-2)

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SUMMARY

The equation of the tangent line to the function y=e^(-2x) at the point (1,e^-2) can be determined by first differentiating the function to find its slope at that point. The derivative f'(x) = -2e^(-2x) leads to f'(1) = -2e^-2, which provides the slope of the tangent line. Using the point-slope form of the equation, y - e^-2 = -2e^-2(x - 1) yields the final equation of the tangent line.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with exponential functions
  • Knowledge of the point-slope form of a line
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study differentiation techniques for exponential functions
  • Learn about the point-slope form of linear equations
  • Explore applications of tangent lines in calculus
  • Review the properties of the exponential function y=e^u
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding tangent lines and their applications in exponential functions.

hvroegindewey
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Does anyone know how to find the equation of a tangent line to y=e^(-2x) at the point (1,e^-2) I honestly have no idea how to even start this problem when I tried it I came up with y=1(x+.27)+ln2
 
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You have to differentiate y=e^(-2x) then solve for f'(1) which will give you the slope. You can solve the equation of the tangent line by plugging values into y-y1=m(x-x1).

Hint: d/du(e^u)=e^u*u'
 

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