determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. 9.) $f(x)=…

Explanation:

Step1: Recall derivative property

The slope of a horizontal tangent line is 0. We need to find the derivative of the function and set it equal to 0.

Step2: Differentiate (f(x)= - 4x+e^{x})

Using the sum - rule and basic derivative formulas, (f^\prime(x)=\frac{d}{dx}(-4x)+\frac{d}{dx}(e^{x})). Since (\frac{d}{dx}(ax)=a) and (\frac{d}{dx}(e^{x}) = e^{x}), we have (f^\prime(x)=-4 + e^{x}).

Step3: Set the derivative equal to 0

Set (f^\prime(x)=0), so (-4 + e^{x}=0). Then (e^{x}=4).

Step4: Solve for (x)

Taking the natural logarithm of both sides, (x = \ln(4)).

Step5: Find the (y) - coordinate

Substitute (x=\ln(4)) into (f(x)): (y=-4\ln(4)+4).

For (g(x)=x + \sin x), (0\leq x<2\pi):

Step1: Differentiate (g(x))

Using the sum - rule and the derivative of (\sin x), (g^\prime(x)=\frac{d}{dx}(x)+\frac{d}{dx}(\sin x)=1+\cos x).

Step2: Set the derivative equal to 0

Set (g^\prime(x)=0), so (1+\cos x = 0). Then (\cos x=-1).

Step3: Solve for (x) in the given interval

In the interval (0\leq x<2\pi), when (\cos x=-1), (x = \pi).

Step4: Find the (y) - coordinate

Substitute (x = \pi) into (g(x)): (y=\pi+\sin(\pi)=\pi).

Answer:

For (f(x)=-4x + e^{x}), the point is ((\ln(4),-4\ln(4)+4)); for (g(x)=x+\sin x), (0\leq x<2\pi), the point is ((\pi,\pi))