1. Suppose that f(x) = x 4 4x 2 4x+7. Draw a graph of y = f(x) including all interesting features.

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1 1. Suppose that f(x) = x 4 4x 4x+7. Draw a graph of y = f(x) including all interesting features. (a) The domain of f is all real numbers. f(0) = 7 and since f(x) = (x 1)(x 3 + x 3x 7) we find that the graph will cross the horizontal axis at (1, 0). f(x) tend to as x tends to ±. (b) f (x) = 4x 3 8x 4 = 4(x + 1)(x x 1). Since x x 1 has two real zeros: x 1 = 1 ( ) 1 5 x = 1 ( ) we see that f (x), being cubic with positive leading coefficient, is first negative, then positive, then negative, then positive, with the sign changes coming (in increasing order) at 1, x 1 and x. Therefore f(x) is Decreasing on (, 1), with a local minimum at ( 1, f( 1)) = ( 1, 8); Increasing on ( 1, x 1 ) with a local maximum at (x 1, f(x 1 )) ( , ); Decreasing on (x 1, x ) with a local minimum at (x, f(x )) ( , ); Increasing on (x, ). There is no absolute maximum, but there is an absolute minimum at (x, f(x )). Also, there must be an additional real zero for some x > x. (c) f (x) = 1x 8 = 4(3x ). f (x) changes from positive to negative to positive with zeros at ± /3 ± Therefore f(x) is convex on (, /3) with an inflection point at ( /3, f( /3)) ( , ); f(x) is concave on ( /3, /3) with an inflection point at ( /3, f( /3)) ( , ); f(x) is convex on ( /3, ).. Suppose that f(x) = (x + 1)/(x + 1). On what intervals is the graph of y = f(x) increasing, decreasing, convex and concave. Sketch the graph and indicate on it each of the points where the graph changes behavior. (a) f(x) is defined for all real numbers and changes sign from negative to positive at x = 1. Its graph has y = 0 as a horizontal asymptote as x tends to ±. (b) Using the quotient rule we see that f (x) = 1 x x (x + 1). The sign of f (x) is the same as that of 1 x x which will change from negative to positive to negative. Therefore 1

2 f(x) is decreasing on (, 1 ) with a local minimum at ( 1, f( 1 )) ( , ); f(x) is increasing on ( 1, 1 + ) with a local maximum at ( 1 +, f( 1 + )) ( , ); f(x) is decreasing on (, 1 +, ). We see that the local maximum is the absolute maximum and the local minimum is the absolute minimum. (c) Using the quotient rule again and taking care to factor out (x + 1) from the numerator and denominator of the resulting expression, we get whose sign is the same as the sign of f (x) = x3 + 3x 3x 1 (x + 1) 3 x 3 + 3x 3x 1 = (x 1)(x + 4x + 1). Since x + 4x + 1 has two real zeros, we see that f (x) will first be negative, then positive, then negative and finally positive. Since the zeros of x +4x+1 are 3 and + 3 we have 3. Suppose that The graph of f(x) is concave on (, 3) with a point of inflection at ( 3, f( 3) ( , ); The graph of f(x) is convex on ( 3, + 3) with a point of inflection at ( + 3, f( + 3) ( , ); The graph of f(x) is concave on ( + 3, 1) with a point of inflection at (1, f(1)) = (1, 1); The graph of f(x) is convex on (1, ). f(x) = 1x 3 ln(x) 4x 3 54x ln(x) + 7x + 7x ln(x) 7x. (a) What is the domain of this function? The domain is all positive real numbers. (b) What is the limit of this function as x approaches 0 from above? Since we know that for p > 0 that we know that lim x 0 xp ln(x) = 0 + lim f(x) = 0. x 0 + (c) Where is this function increasing and where is it decreasing? By applying the product rule and factoring we have f (x) = 36(x 1)(x ) ln(x). Since both ln(x) and (x 1) change from negative to positive at x = 1, f (x) changes from positive to negative only at x =. Therefore

3 f(x) is decreasing on (0, ) with a local minimum at (, f()) (, ). (d) Sketch the graph of this function, indicating all places where it changes from increasing to decreasing and from decreasing to increasing. Observe also that there is a removable discontinuity at (0, 0) and that as x tends to infinity so does f(x). 4. Suppose that f(x) = exp( x /). Let h(x) = f (x). Draw a graph of y = h(x) indicating where the function changes from increasing to decreasing, decreasing to increasing, convex to concave and concave to convex. We showed on an earlier assignment that h(x) = (x 1) exp( x /). This is the function we wish to graph. (a) First we see that the graph is symmetric with respect to the vertical axis. f(x) > 0 for x > 1 and f(x) < 0 for x < 1. The points (1, 0), ( 1, 0) and (0, 1) are all on the graph. y = 0 is a horizontal asymptote as x approaches ± since we showed that for p, q > 0 that lim x x p exp(x q ) = 0. (b) By using the product rule we see that h (x) = x(3 x ) exp(x) which changes sign at 0 and ± 3. We see that h(x) is increasing on (, 3) with a local maximum at ( 3, h( 3)) ( , ) h(x) is decreasing on ( 3, 0) with a local minimum at (0, h(0)) = (0, 1). h(x) is increasing on (0, 3) with a local maximum at ( 3, h( 3)) ( , ) h(x) is decreasing on ( 3, ). The local maxima are the absolute maxima and the local minimum is the absolute minimum. (c) Applying the product rule again we get h (x) = (x 4 6x + 3) exp( x /). h (x) has the same sign as x 4 6x + 3 which has four real zeros: 3 + 6, 3 + 6, 3 6, 3 6 which are approximately , , , We find these by letting u = x in x 4 6x + 3 = 0 and solving u 6u + 3 = 0 first. Therefore we have h(x) is convex on (, 3 + 6) with an inflection point approximately at ( , ); h(x) is concave on ( 3 + 6, 3 6) with an inflection point approximately at ( , ); h(x) is convex on ( 3 6, 3 6) with an inflection point approximately at ( , ); 3

4 h(x) is convex on ( 3 6, 3 + 6) with an inflection point approximately at ( , ); h(x) is concave on ( 3 + 6, ). 5. Hermite polynomials. Let f(x) be as in the preceding problem. We can define a sequence of polynomials, H n (x), according to the rule For example, H n (x) = ( 1)n f(x) d n dx n f(x). H 0 (x) = ( 1) 0 1 f(x) f(x) = 1 H 1 (x) = 1 f(x) f (x) = exp(x /)(exp( x /)( x)) = x Find and graph H 3 (x) and H 4 (x). H (x) = ( 1) f(x) f (x) = exp(x /)(exp( x /)(x 1)) = x 1. From our calculations in the previous problem we see that H 3 (x) = 1 d 3 exp( x ) dx 3 exp( x /) = 1 exp( x ) h (x) = x(x 3) H 4 (x) = 1 d 4 exp( x ) dx 4 exp( x /) = 1 exp( x ) h (x) = x 4 6x + 3. Turning first to H 3 (x), we know that it is 0 at 0 and ± 3. H 3(x) = 3x 3 = 3(x 1) so H 3 (x) is increasing on (, 1) and (1, ) and decreasing on ( 1, 1). H 3 (x) = 6x so H 3 (x) is concave on ( ) and convex on (0, ). Now we look at H 4 (x). We found its zeros above. We have H 4(x) = 4x 3 1x = 4x(x 3) so H 4 (x) is decreasing on (, 3) and on (0, 3) while it is increasing on ( 3, 0) and on ( 3, ). The local minima are at (± 3, 6) and the local maximum is at (0, 3). For convexity and concavity we have H 4 (x) = 1(x 1) so the graph is convex on x > 1 and concave on x < 1. The points of inflection are (±1, ). 4

5 6. Go to and open the link Finding Rational Zeros of Polynomials. Go to page 310 and read problem 89. Find the value of x that gives the open box of maximum volume by sketching the graph of the volume as a function of x. Note that you do not need to consider where the graph is convex or concave. If we let V (x) denote the volume of the box (in cubic inches) if squares of side length x inches are removed, then V (x) = x(5 x)(8 x) = 4x 3 6x + 40x = (x 3 13x + 0x) for 0 x 5/. V (0) = V (5/) = 0 and V (x) > 0 if 0 < x < 5/. V (x) = (6x 6x + 0) = 4(3x 13x + 10) = 4(x 1)(3x 10) The factor (3x 10) is always negative (for 0 x 5/ is the domain), so V (x) changes from positive to negative at x = 1. This tells us the the maximum value of V (x) occurs at x = 1. We should cut out squares that are one inch by one inch. 5