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Calculus BC: Applications of the Derivative

Problems

Analysis of Graphs

Optimization

Problem : Find the critical points and inflection points of the function f (x) = x 4 -2x 2 (with domain the set of all real numbers). Which of the critical points are local minima? local maxima? Is there a global minimum or maximum?

We first calculate the derivatives of the function:


f'(x) = 4x 3 - 4x  
  = 4(x + 1)x(x - 1)  
f''(x) = 12x 2 - 4  
  = 4(3x 2 - 1)  

We see that f'(x) = 0 when x = - 1 , 0 , or 1 , so these are the three critical points of f . We calculate the second derivatives at these points:


f''(- 1) = 8  
f''(0) = -4  
f''(1) = 8  

so by the second derivative test, f has local minima at -1 and 1 and a local maximum at 0 . Substituting back into the original function yields


f (- 1) = -1  
f (0) = 0  
f (1) = -1  

so f attains its global minimum of -1 at x = ±1 . It is clear from the graph of f that it has no global maximum.
Figure %: Graph of f (x) = x 4 -2x 2
To find the points of inflection, we solve f''(x) = 0 , or 12x 2 - 4 = 0 , which has solutions x = ±1/3) ±0.58 . Referring once again to the graph of f , we can check that the concavity does indeed change at these x -values.

Problem : Find the inflection points of f (x) = e -x2 . (This famous function is called the gaussian.)

We compute the derivatives:


f'(x) = -2xe -x2 ,  
f''(x) = (- 2x)(- 2xe -x2 ) + (- 2)(e -x2 )  
  = (4x 2 -2)e -x2  

Solving f''(x) = 0 for x , we get the inflection points x = ±1/ 0.71 . This is reasonable when one considers the graph of f .
Figure %: Graph of f (x) = e-x2

Problem : Find a function f (x) with inflection points at x = 1 and x = 2 .

We need f''(1) = f''(2) = 0 , so we might as well let

f''(x) = (x - 1)(x - 2) = x 2 - 3x + 2    

We search for a function with this derivative, ultimately obtaining

f'(x) = - + 2x    

Similarly, a function with this derivative is given by

f (x) = - + x 2    

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