Problem 124 Graph cach function. Then estima... [FREE SOLUTION] (2024)

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To find the first derivative of a function, you perform differentiation. For the function \(f(x) = x^{2}(x-2)^{3}\), you need to use both the product rule and the chain rule. The product rule states that if you have two functions multiplied together, \(u(x)\) and \(v(x)\), then the derivative is given by \(u'(x)v(x) + u(x)v'(x)\). Applying this to our function, let \(u(x) = x^{2}\) and \(v(x) = (x-2)^{3}\).
The derivatives are \(u'(x) = 2x\) and \(v'(x) = 3(x-2)^{2}\). Combining these using the product rule gives us:
\[ f'(x) = 2x(x-2)^{3} + x^{2} \times 3(x-2)^{2} \] This expression can then be simplified further. Tips for finding and simplifying first derivatives:

• Identify and clearly separate the parts of the product.
• Apply the chain rule when there is a composite function involved.
• Simplify your results by factoring out common terms where possible.

Relative extrema refer to points in the domain of a function where the function reaches a relative maximum or minimum compared to nearby points. To find the relative extrema of \(f(x)\), you first need to find the critical points by setting the first derivative equal to zero and solving for \(x\).
These points are where potential relative extrema can occur. After finding the critical points, determine if they are relative maxima or minima by checking the second derivative or the behavior of the first derivative around these points. It involves determining the change in sign of the first derivative:

• If \(f'(x)\) changes from negative to positive, the point is a relative minimum.
• If \(f'(x)\) changes from positive to negative, the point is a relative maximum.

The second derivative test helps in determining the concavity of the function and the nature of the critical points. The test involves taking the second derivative of the function and evaluating it at the critical points.
For the function \(f(x) = x^{2}(x-2)^{3}\), compute \(f''(x)\) to use this test:

• If \(f''(x) > 0\), the function is concave up at \(x\) and has a relative minimum.
• If \(f''(x) < 0\), the function is concave down at \(x\) and has a relative maximum.
• If \(f''(x) = 0\), the test is inconclusive. You may need to use the first derivative test again.

Applying this to our function and computing \(f''(x)\) can be complex, but it confirms the relative extrema found through sign analysis.

The product rule is an essential calculus tool used to differentiate products of two functions. For a product of two functions \(u(x)\) and \(v(x)\), the derivative is:
\[ (uv)' = u'v + uv' \] In the case of \( f(x) = x^{2}(x-2)^{3} \), let \(u(x) = x^{2} \) and \(v(x) = (x-2)^{3}\). Following the product rule:
\[ f'(x) = 2x(x-2)^{3} + x^{2} \times 3(x-2)^{2} \] Tips for applying the product rule:

• Identify and rewrite the function as a product of simpler functions.
• Apply the product rule step-by-step, differentiating each part individually.
• Combine and simplify the resulting expression.

Consistent practice with the product rule will improve your differentiation skills.

The chain rule is crucial for differentiating composite functions, where one function is nested inside another. The rule states that if you have \(g(h(x))\), the derivative is given by \(g'(h(x)) \times h'(x)\).
For the function \(f(x) = (x-2)^{3}\), let \(u(x) = x-2 \) and \(v(u) = u^{3} \). Then, \(f(x) = v(u(x))\). Applying the chain rule:
\[ \frac{d}{dx}[u^{3}] = 3u^{2} \times u'(x) = 3(x-2)^{2} \] This derivative is used in the product rule calculation for \(f'(x)\). Key points for the chain rule:

• Identify the inner and outer functions.
• Differentiate the outer function, keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

Understanding and applying the chain rule simplifies many complex derivatives.