Integrated Calculus I Quiz 4 Solutions, 3/26/4. Question 1. Let f(x) = x3 − 15x2 +
48x − 15, defined for all real x. • Find the intervals on which f is increasing.
Integrated Calculus I Quiz 4 Solutions, 3/26/4 Question 1 Let f (x) = x3 − 15x2 + 48x − 15, defined for all real x. • Find the intervals on which f is increasing. We have f 0 = 3x2 − 30x + 48 = 3(x2 − 10x + 16) = 3(x − 2)(x − 8). So f is increasing on the intervals where the one of the factors x − 2 and x − 8 is zero or both factors have the same sign, so for x ≥ 8 or x ≤ 2, so on the intervals (−∞, 2] and [8, ∞). • Find the intervals on which f is decreasing. The function f is decreasing on the intervals where one of the factors x − 2 and x − 8 is zero or the factors have opposite signs, so when 2 ≤ x ≤ 8, so on the interval [2, 8]. • Find the local maxima and minima of the function f . We have critical points at x = 2 and x = 8 only. We have f 00 = 6x − 30 = 6(x − 5). – When x = 2, f 00 < 0, so x = 2 is a local maximum. Then we have: y = f (2) = 23 − 15(22 ) + 48(2) − 15 = 8 − 60 + 96 − 15 = 29. So (2, 29) is a local maximum. – When x = 8, f 00 > 0, so x = 8 is a local maximum. Then we have: y = f (8) = 83 − 15(82 ) + 48(8) − 15 = 512 − 960 + 384 − 15 = −79. So (8, −79) is a local minimum. – Since f is differentiable, there are no other local extrema.
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• Find the absolute maxima and minima of the function f on the interval [0, 10], explaining your argument. – Putting x = 0, we get f (0) = −15.
– Next, putting x = 10, we get: f (10) = 103 −15(102 )+48(10)−15 = 1000−1500+480−15 = −35. – Since f is differentiable, the absolute minimum of y = f (x) occurs at an endpoint of the interval of definition, or at a critical point. So here the absolute minimum is at x = 8 and is y = −79. – Since f is differentiable, the absolute maximum of y = f (x) occurs at an endpoint of the interval of definition, or at a critical point. So here the absolute maximum is at x = 2 and is y = 29.
• Find the intervals on which the graph of f is concave up and the intervals on which the graph of f is concave down. Since f 00 (x) = 6(x − 5), we have f 00 (x) ≥ 0, for x ≥ 5 and f 00 (x) ≤ 0, for x ≤ 5, so the graph is concave up on the interval [5, ∞) and concave down on the interval (−∞, 5]. • Explain why we know that f has a root in the interval [0, 1] and use Newton’s method with three iterations to estimate the root. – Putting x = 0, we find: f (0) = −15.
– Putting x = 1, we find: f (1) = 13 − 15(1)2 + 48(1) − 15 = 1 − 15 + 48 − 15 = 19.
– So the function f changes sign on the interval [0, 1]. But, since f is a polynomial, it is continuous. So by the Intermediate Value Theorem, it therefore takes the value 0 somewhere in the open interval (0, 1).
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The iteration map for Newton’s formula is: x → N (x) = x −
x3 − 15x2 + 48x − 15 f (x) = x − f 0 (x) 3x2 − 30x + 48
3x3 − 30x2 + 48x − (x3 − 15x2 + 48x − 15) = 3x2 − 30x + 48
2x3 − 15x2 + 15 . 3x2 − 30x + 48 Choosing x = 0, for our initial point x0 , we get: =
x1 = N (x0 ) = N (0) = x2 = N (x1 ) =
5 15 = = 0.3125, 48 16
250 163 75 162
− −
375 162 150 16
+ 15 + 48
50 − 75(16) + 3(16)3 5 = 48 25 − 50(16) + 163 5 50 − 1200 + 12288 = 48 25 − 800 + 4096 27845 5 5569 = = = 0.3493551, 24 3321 79704
3355637817404105 = 0.3498580. 9591428768976036 So our estimate for the root is x3 = 0.3498580. In fact we have: x3 = N (x2 ) =
f (x3 ) =
−3112859423230357561914622780498504638671875 882368341237636829514131869819624542795986734656 = −3.527845773(10−6).
So our estimate gives f quite close to zero. • Sketch the graph of the function f on the interval [0, 10]. See the Maple graphics for this.
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Question 2 √ 3 Determine the linear approximation to the function f (x) = x based at √ 3 x = 8 and use your approximation to estimate 8.1. Also sketch the graphs of the function f and of its linear approximation √ on the interval [6, 12] and use the graphs to decide if your estimate of 3 8.1 is an under-estimate or over-estimate, explaining your answer. The linear approximation based at x = a of a function f (x) is given by the formula: flin (x) = f (a) + f 0 (a)(x − a). 1 2 Here we have f (x) = x 3 , so f 0 (x) = 31 x− 3 . 1 Next, a = 8, so f (a) = f (8) = 8 3 = 2. 2 1 Also f 0 (a) = f 0 (8) = 13 8− 3 = 31 2−2 = 12 . So the required linear approximation is: flin (x) = f (8) + f 0 (8)(x − 8) = 2 +
1 1 (x − 8) = (x + 16). 12 12
The graphs of f and flin (x) show that they remain very close over the interval [6, 12]. √ We may then estimate 3 8.1 = f (8.1), by: flin (8.1) =
241 1 (8.1 + 16) = = 2.00833. 12 120
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We have f 00 (x) = − 29 x− 3 < 0 near x = 8, so the graph of f (x) is concave down near x = 8. So the tangent line at x = 8, which is the graph of the linear approximation, flin (x), lies slightly above the graph of f (x). So the estimate flin (8.1) is a slight overestimate of f (8.1). √ 3 Maple gives 8.1 = 2.00829885, so the linear approximation is correct to one five-hundredth of one percent.
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