Question

the derivative of a function ( f ) is given by ( f(x)=0.1x + e^{0.25x} ). at what value of ( x ) for ( x>0 ) does the line tangent to the graph of ( f ) at ( x ) have slope 2?
a 0.512
b 1.849
c 2.287
d 8.113

Explanation:

Step1: Set up the equation

The slope of the tangent line to the graph of \(y = f(x)\) at \(x\) is given by \(f^{\prime}(x)\). We want to find \(x>0\) such that \(f^{\prime}(x)=2\). So, we set up the equation \(0.1x + e^{0.25x}=2\).

Step2: Use a numerical method (e.g., Newton - Raphson or a graphing utility)

Let \(g(x)=0.1x + e^{0.25x}-2\).
If we use a graphing calculator:

• Enter the function \(y = 0.1x+e^{0.25x}\) and \(y = 2\).
• Use the intersection feature.

When we input \(x = 1.849\):
\(0.1\times1.849+e^{0.25\times1.849}=0.1849 + e^{0.46225}\)
\(e^{0.46225}\approx1.588\), and \(0.1849+1.588 = 1.7729\) (not correct).
When \(x = 2.287\):
\(0.1\times2.287+e^{0.25\times2.287}=0.2287+e^{0.57175}\)
\(e^{0.57175}\approx1.771\), and \(0.2287 + 1.771=2.0\) (approximately).

Answer:

C. \(2.287\)