Find the intervals on which the graph of \( f \) is concave upward, the intervals on which the graph of \( f \) is concave downward, and the inflection points \( f(x)=-x^{6}+6 x^{5}-6 x+7 \)

I apologize for the inconvenience. Let's try to find the intervals on which the graph of \( f \) is concave upward, the intervals on which the graph of \( f \) is concave downward, and the inflection points of the function \( f(x)=-x^{6}+6x^{5}-6x+7 \) manually. To find the intervals on which the graph of \( f \) is concave upward, we need to determine where the second derivative of \( f \) is positive. Similarly, to find the intervals on which the graph of \( f \) is concave downward, we need to determine where the second derivative of \( f \) is negative. The inflection points occur where the concavity changes, which is where the second derivative is equal to zero. Let's start by finding the second derivative of \( f(x)=-x^{6}+6x^{5}-6x+7 \) and then determine the intervals of concavity and inflection points. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(\frac{d}{dx}\left(-x^{6}+6x^{5}-6x+7\right)\right)\) - step1: Calculate: \(\frac{d}{dx}\left(-6x^{5}+30x^{4}-6\right)\) - step2: Use differentiation rules: \(-\frac{d}{dx}\left(6x^{5}\right)+\frac{d}{dx}\left(30x^{4}\right)-\frac{d}{dx}\left(6\right)\) - step3: Calculate: \(-30x^{4}+120x^{3}-0\) - step4: Remove 0: \(-30x^{4}+120x^{3}\) The second derivative of the function \( f(x)=-x^{6}+6x^{5}-6x+7 \) is \( -30x^{4}+120x^{3} \). To find the intervals on which the graph of \( f \) is concave upward, we need to determine where the second derivative is positive. To find the intervals on which the graph of \( f \) is concave downward, we need to determine where the second derivative is negative. The inflection points occur where the concavity changes, which is where the second derivative is equal to zero. Let's find the critical points of the second derivative by setting it equal to zero and solving for \( x \) to determine the inflection points. Solve the equation by following steps: - step0: Solve for \(x\): \(-30x^{4}+120x^{3}=0\) - step1: Factor the expression: \(30x^{3}\left(-x+4\right)=0\) - step2: Divide both sides: \(x^{3}\left(-x+4\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x^{3}=0\\&-x+4=0\end{align}\) - step4: Set the base equal to 0: \(\begin{align}&x=0\\&x=4\end{align}\) - step5: Rewrite: \(x_{1}=0,x_{2}=4\) The critical points of the second derivative are \( x=0 \) and \( x=4 \). Now, we need to determine the intervals of concavity and inflection points by analyzing the sign of the second derivative in the intervals around these critical points. 1. For \( x<0 \): - The second derivative is negative, so the graph of \( f \) is concave downward in this interval. 2. For \( 04 \): - The second derivative is negative, so the graph of \( f \) is concave downward in this interval. Therefore, the intervals on which the graph of \( f \) is concave upward are \( 04 \). The inflection points occur at \( x=0 \) and \( x=4 \).