Tangent to Curve $e^x+k$ at $x=a$: Find $k$

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SUMMARY

The discussion focuses on determining the value of \( k \) in the function \( f(x) = e^x + k \) such that the tangent at \( x = a \) intersects the origin. The tangent line is expressed as \( y - f(a) = f'(a)(x - a) \). By substituting \( x = 0 \) and \( y = 0 \) into the tangent equation, the solution reveals that \( k = e^a(a - 2) \). This establishes a clear relationship between \( k \) and the point of tangency \( a \).

PREREQUISITES
  • Understanding of exponential functions, specifically \( e^x \)
  • Knowledge of derivatives and tangent lines in calculus
  • Familiarity with solving equations involving variables
  • Basic algebraic manipulation skills
NEXT STEPS
  • Explore the properties of exponential functions and their derivatives
  • Learn about tangent lines and their equations in calculus
  • Study applications of derivatives in real-world problems
  • Investigate the implications of changing the value of \( a \) on the function and tangent
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Students and educators in calculus, mathematicians interested in function analysis, and anyone studying the properties of exponential functions and their tangents.

Bushy
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Hi there,

The function $f(x)= e^x+k$ has a tangent to the curve at $x=a$ and going through the origin. Find $k$ in terms of $a$
 
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Bushy said:
The function $f(x)= e^x+k$ has a tangent to the curve at $x=a$ and going through the origin. Find $k$ in terms of $a$
The equation of the tangent line is $y-f(a)=f^{\prime}(a)(x-a)$.
Now let $x=0~\&~y=0$ then solve for $k$.
 
$y-f(a) = f'(a)(x-a)$

and

$0-(e^a+k) = e^a(0-a)$

so

$k=e^a(a-2)$
 

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