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Problem 975

# Determine the critical points of $\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}-4 \mathrm{x}^{3}$ and sketch the graph.

Expert verified
The critical points of the function $$f(x) = 3x^4 - 4x^3$$ are $$x = 0$$ (inflection point) and $$x = 1$$ (local minimum). The graph starts at a positive value, decreases to a local minimum at $$x = 1$$, and then increases indefinitely.
See the step by step solution

## Step 1: Find the derivative of $$f(x)$$

We need to find the derivative of the function $$f(x) = 3x^4 - 4x^3$$. To do this, take the derivative with respect to $$x$$: $f'(x) = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2$

## Step 2: Set the derivative equal to zero and solve for $$x$$

Now, we need to find the critical points by setting the derivative equal to zero: $12x^3 - 12x^2 = 0$ We can factor out a $$12x^2$$: $12x^2(x - 1) = 0$ This gives us two critical points: $x = 0, \quad x = 1$

## Step 3: Determine the nature of the critical points

Now we'll use the second derivative test to determine whether the critical points are local minima, local maxima, or saddle points. First, find the second derivative: $f''(x) = \frac{d^2}{dx^2}(12x^3 - 12x^2) = 36x^2 - 24x$ Now, evaluate the second derivative at the critical points: $f''(0) = -24(0) = 0$ $f''(1) = 36(1) - 24(1) = 12$ Since $$f''(0) = 0$$, we cannot determine the nature of the critical point at $$x = 0$$ using the second derivative test. However, since $$f''(1) > 0$$, we can conclude that there is a local minimum at $$x = 1$$.

## Step 4: Sketch the graph of $$f(x)$$

Now, using the critical points and the nature of the critical points, we can sketch the graph of $$f(x) = 3x^4 - 4x^3$$. At $$x = 0$$, the function has an inflection point, and at $$x = 1$$, we have a local minimum. The graph starts at a positive value, decreases to a local minimum at $$x = 1$$, and then increases.

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