Slope of line tangent to the graph f(x) at x=3 is [tex]10e^9[/tex]
Given :
Function [tex]f(x)=xe^{3x}[/tex]
To find slope of the line tangent to the graph , we take derivative
Lets take derivative of given f(x)
Apply product rule to take derivative
[tex]f(x)=\left(xe^{3x}\right)\\f'(x)=1\cdot \:e^{3x}+e^{3x}\cdot \:3x\\f'(x)=e^{3x}+3e^{3x}x\\f'(x)=e^{3x}(1+3x)[/tex]
Now we find out slope of tangent line when x=3
Substitute 3 for x in f'(x)
[tex]f'(x)=e^{3x}(1+3x)\\x=3\\f'(3)=e^{3(3)}(1+3(3))\\\\f'(3)=e^{9}(10)\\\\[/tex]
Slope of line tangent to the graph at x=3 is [tex]10e^9[/tex]
Learn more : brainly.com/question/12106202
