Help Let f(x) = 10-6x2 - x3. A portion of the graph of f (generated using MAPLE

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Let f(x) = 10-6x2 - x3. A portion of the graph of f (generated using MAPLE software) is provided as the last page of this worksheet for your convenience as you consider the exercises given below. You may use this page as part of your final workshop report or reproduce your results on another sheet. Additional copies of page 3 are also available from the workshop coordinator. Compute and simplify the derivative f'(x) by applying the differentiation rules found in section 3.3 of Stewart. Determine the slope-intercept equation of the line tangent to the graph of f at the point (1, f (1)). Label this point on your curve, then sketch and label this tangent line. Repeat part b) for the point (-3, f (-3)). Draw the line segment joining the points (1, f(1)) and (-3, f(-3)). Determine the slope of this line, then locate all points on the graph of f where the tangent line has this same slope. Plot and label each such point and sketch its tangent line. (Your coordinates should be accurate to at least three decimal places.) Carefully sketch the graph of f' on a new set of axes. Plot the points on this curve corresponding to x = 1 and x = -3 and briefly explain how these points are related to their counterparts on the graph of f. (Refer back to parts b) and c) for your answer.) Next locate and label the x-intercepts of this new curve. What significance do these points have relative to the graph of f ? The line described in part b) intersects the graph of f both at (1, f(1)) and at a second point. Find both coordinates of this second intersection point.

Explanation / Answer

f(x) = 10 - 6x^2 -x^3

then df(x)/dx = 0 - 6 * 2x - 3x^2 = -12x - 3x^2 = -3x(4 + x)

then slope at x = 1 is df(x)/dx at x = 1 = -3*1 ( 5) = -15

So the slope will be 15 units.

The equation of the line is y = mx + c where m is the slope and c is the y-intercept

Now at x = 1 f(x) = 10 - 6 -1 = 3

so at point(1,3) c = 3 - (-15*1) = 3 + 15 = 18

so y = 15x + 18 which is the answer to part b

for part c, at x = 3, slope = -3 * 3 (7) = -63

and f(-3) = 10 - 6(9) - 27 = -71

then c = y - mx = -71- (-63 * -3) = -71 - 189 = -260

so the equation is y = -63x -260

part d) f(1) = 3 and f(-3) = -71

then line between these point will have a slope of f(-3) -f(1) / (-3 - 1) = 18.5

the y = 18.5 x + c where the line passes through both these points

then c = 3 - 18.5 * 1 = 15.5

so the equation of the line is y = 18.5x + 15.5

now -3x ( 4 + x) = 18.5 then this will be a quadratic eqation
6x^2 + 24x +37 = 0

The roots of this equation turn out to be imagenary, hence there will not be any such points

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