the derivative of a function ( f ) is given by ( f(x)=0.1x + e^{0.25x} ). at what value of ( x ) for ( x>0 )…

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?\na 0.512\nb 1.849\nc 2.287\nd 8.113

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?\na 0.512\nb 1.849\nc 2.287\nd 8.113

Answer

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)