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Magazine Chapter 4: Problem 4
Determine the value of the second derivative at the value indicated. State whether the curve lies above or below the tangent at this point. a. \(f(x)=2 x^{3}-10 x+3\) at \(x=2\) b. \(g(x)=x^{2}-\frac{1}{x}\) at \(x=-1\) c. \(p(w)=\frac{w^{2}}{\sqrt{w^{2}+1}}\) at \(w=3\) d. \(s(t)=\frac{2 t}{t-4}\) at \(t=-2\)
Short Answer
Expert verified
a: 24, above; b: 4, above; c: negative, below; d: negative, below.
Step by step solution
01
Find the first derivative of f(x)
To begin solving part a, calculate the first derivative of the function \(f(x) = 2x^3 - 10x + 3\). The first derivative is obtained by applying basic differentiation rules to each term: \(f'(x) = 6x^2 - 10\).
02
Calculate the second derivative of f(x)
Next, differentiate \(f'(x)\) to find the second derivative \(f''(x)\). Apply the power rule: \(f''(x) = 12x\).
03
Evaluate the second derivative at x=2
Substitute \(x = 2\) into the second derivative to find its value: \(f''(2) = 12(2) = 24\). Since \(f''(2) > 0\), the curve is concave up at this point.
04
Find the first derivative of g(x)
For part b, differentiate \(g(x) = x^2 - \frac{1}{x}\) to find the first derivative: \(g'(x) = 2x + \frac{1}{x^2}\).
05
Calculate the second derivative of g(x)
Differentiate \(g'(x)\) to get \(g''(x) = 2 - \frac{2}{x^3}\).
06
Evaluate the second derivative at x=-1
Substitute \(x = -1\) into \(g''(x)\): \(g''(-1) = 2 - \frac{2}{(-1)^3} = 2 + 2 = 4\). Since \(g''(-1) > 0\), the curve is concave up.
07
Find the first derivative of p(w)
For part c, use the quotient rule to find the first derivative of \(p(w) = \frac{w^2}{\sqrt{w^2+1}}\): \(p'(w) = \frac{w(w^2+1)^{1/2} - w^2(1/2)(w^2+1)^{-1/2}(2w)}{(w^2+1)}\).
08
Simplify and find the second derivative of p(w)
Simplify further and use the quotient rule or software to derive \(p''(w)\). Calculating directly, \(p''(w) = \frac{-w^4 + 2w^2 - 1}{(w^2+1)^{3/2}}\).
09
Evaluate the second derivative at w=3
Substitute \(w = 3\) into \(p''(w)\): \(p''(3) = \frac{-3^4 + 2 \times 3^2 - 1}{(9+1)^{3/2}} = \frac{-81 + 18 - 1}{1000^{3/2}} = \frac{-64}{1000}\). Since \(p''(3) < 0\), the curve is concave down.
10
Find the first derivative of s(t)
For part d, use the quotient rule: \(s(t) = \frac{2t}{t-4}\). First derivative: \(s'(t) = \frac{2(t-4)-2t}{(t-4)^2} = \frac{-8}{(t-4)^2}\).
11
Calculate the second derivative of s(t)
Differentiate \(s'(t)\) using the chain rule: \(s''(t) = \frac{16(t-4)}{(t-4)^4}\).
12
Evaluate the second derivative at t=-2
Substitute \(t = -2\) into \(s''(t)\): \(s''(-2) = \frac{16(-6)}{256} = \frac{-96}{256} = -0.375\). Since \(s''(-2) < 0\), the curve is concave down.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Concavity
Concavity is a concept in calculus that helps us understand the shape and behavior of a curve on a graph. It tells us whether a curve is curving upwards or downwards at a particular point. This is determined by looking at the second derivative of a function.
- If the second derivative is positive (\(f''(x) > 0\)), the curve is concave up. This means it looks like a U-shape and lies above the tangent line at that point.
- If the second derivative is negative (\(f''(x) < 0\)), the curve is concave down, resembling an upside-down U, and lies below the tangent line.
Differentiation
Differentiation involves finding the derivative of a function, which essentially calculates how a function's output changes with respect to changes in its input. It is a foundational concept in calculus that helps us understand the rate of change and slopes of curves.
- The first derivative (\(f'(x)\)) tells us about the slope or steepness of the curve at any given point.
- For the second derivative (\(f''(x)\)), it goes a step further to provide insights into the acceleration or the change of the slope itself.
Quotient Rule
The quotient rule is a nifty tool in calculus used to differentiate functions that are the ratio of two other functions. It's essential when dealing with fractions or rational functions. The quotient rule states:
If you have a function \( h(x) = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are differentiable, then the derivative is given by:
\[ h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
If you have a function \( h(x) = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are differentiable, then the derivative is given by:
\[ h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
- This rule helps calculate the rate of change for ratios of functions accurately.
- Apply it when you encounter a function that requires division.
Power Rule
The power rule is one of the simplest and most used differentiation techniques in calculus. It applies to functions where the variable is raised to a power, such as polynomials. The power rule states:
If \( f(x) = x^n \), where \( n \) is a real number, then the derivative is:
\[ f'(x) = nx^{n-1} \]
If \( f(x) = x^n \), where \( n \) is a real number, then the derivative is:
\[ f'(x) = nx^{n-1} \]
- This rule makes it easy to find derivatives for polynomial terms by simply multiplying the power by the coefficient and reducing the power by one.
- It's a crucial rule for quickly differentiating terms like \( x^2 \), \( x^3 \), or even \( 5x^4 \).
