Question

determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. 9.) $f(x)= - 4x + e^{x}$ 10.) $g(x)=x+sin x, 0leq x<2pi$

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 \((\pi,\pi)\)