where does the function 4e^x-x have a horizontal tangent line ?

Question

anonymous · Answer

i have a review sheet it says to put dy/dx =0 , so then 4e^x-1 it says to solve for x. the answer is ln(1/4)=-1.39 but i dont understand how my professor got this answer?

anonymous · Answer

So where is 4e^x -1 = 0 ?

stamp · Answer

$$f(x)=4e^x-x$$A function graphs points to form a line. The derivative of the function tells you the slope of the function at certain points. A slope of zero is also a flat line. $$f'(x)=4e^x-1$$This is the derivative of f(x). So at any point f'(x) we have an output in the form of the slope of the line.

When the slope equals zero, the tangent line is flat. The slope is f'(x), so we want to know what values of x will f'(x)=0.
$$f'(x)=4e^x-1=0
$$Solve for x.$$4e^x-1=0$$$$4e^x=1$$$$e^x=1/4$$Natural log of both sides.$$lne^x=ln1/4$$$$x=ln1/4$$