Calculus - Tangent Line Question

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SUMMARY

The problem involves finding the value of b for which the line y=10x is tangent to the curve y=e^{bx}. The solution requires setting up a system of equations: 10x=e^{bx} and 10=be^{bx}. By manipulating these equations, it is determined that x must equal 1/b, leading to the conclusion that x=10/e. This establishes that the correct value of b is 10.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and tangent lines.
  • Familiarity with exponential functions and their properties.
  • Ability to solve systems of equations.
  • Basic knowledge of GRE mathematics problems.
NEXT STEPS
  • Study the properties of exponential functions, particularly e^{bx}.
  • Learn how to find derivatives of functions and apply them to tangent line problems.
  • Practice solving systems of equations involving exponential and linear functions.
  • Review GRE mathematics problem-solving strategies for efficiency.
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Students preparing for the GRE subject test, educators teaching calculus concepts, and anyone looking to improve their problem-solving skills in mathematics.

rakalakalili
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Homework Statement


Hello, this is a problem from the practice test for the GRE subject test.
For what value of b is the line y=10x tangent to the curve y=e^{bx} at some point in the xy-plane?
A) \frac{10}{e}
B)10
C)10e
D)e^{10}
E)e

Homework Equations


The Attempt at a Solution


For the line to be tangent to the curve, they must intersect at a point and their derivatives must be equal. So we get a system of equations:
10x=e^{bx}
10=be^{bx}
But I cannot think of a way to solve this system easily. Would the best method be to simply plug in the possible answers and check? This seems time consuming, and since this is a practice GRE question I am interested in the simplest way of solving it. Please let me know if I am over looking something, thank you!
 
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Hint:
e^bx= 1+bx+o(x^2)
and we have that (e^bx-1)/x=b+o(x)

So because the derivative of e^bx should be equal the slope of the tangent line, we concur?

I just picked the point zero randomly.
 

You have to solve two equations for x and for b:
10x = e^(bx) and 10 = b*e^(bx)

Multiplying the second eqn by x and comparing it to the first eqn gives

e^bx = 10x = x*b*e^(bx)

It follows that 1 = x*b, so x = 1/b. A little more work will reveal that x = e/10.
 

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