Answer:
Step-by-step explanation:
Our friend asking what the actual function is has a point. I completed this under the assumption that what we have is:
[tex]f(x)=\frac{e^{3x}}{5x-2}[/tex] and used the quotient rule to find the derivative, as follows:
[tex]f'(x)=\frac{e^{3x}(5)-[(5x-2)(3e^{3x})]}{(5x-2)^2}[/tex] and simplifying a bit:
[tex]f'(x)=\frac{5e^{3x}-[15xe^{3x}-6e^{3x}]}{(5x-2)^2}[/tex]and a bit more to:
[tex]f'(x)=\frac{5e^{3x}-15xe^{3x}+6e^{3x}}{(5x-2)^2}[/tex] and combining like terms:
[tex]f'(x)=\frac{11e^{3x}-15xe^{3x}}{(5x-2)^2}[/tex] and factor out the GFC in the numerator to get:
[tex]f'(x)=\frac{e^{3x}(11-15x)}{(5x-2)^2}[/tex] That's the derivative simplified. If we want f'(0), we sub in 0's for the x's in there and get the value of the derivative at x = 0:
[tex]f'(0)=\frac{e^0(11-15(0))}{(5(0)-2)^2}[/tex] which simplifies to
[tex]f'(0)=\frac{11}{4}[/tex] which translates to
The slope of the function is 11/4 at the point (0, -1/2)
