The graph of $g(x) = ax^3 + bx^2 + cx$, a cubic function having a y-intercept of 0, is drawn below - NSC Mathematics - Question 8 - 2020 - Paper 1
Question 8
The graph of $g(x) = ax^3 + bx^2 + cx$, a cubic function having a y-intercept of 0, is drawn below. The x-coordinates of the turning points of $g$ are -1 and 2.
8.1... show full transcript
Worked Solution & Example Answer:The graph of $g(x) = ax^3 + bx^2 + cx$, a cubic function having a y-intercept of 0, is drawn below - NSC Mathematics - Question 8 - 2020 - Paper 1
Step 1
For which values of $x$ will $g$ increase?
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Answer
To determine for which values of x the function g increases, we need to look at the first derivative g′. A function increases when its derivative is positive. Given the turning points at x=−1 and x=2, we analyze the intervals:
For x<−1, g′>0 (increasing).
For −1<x<2, g′<0 (decreasing).
For x>2, g′>0 (increasing).
Thus, g increases for xextin(−ext∞,−1)extand(2,ext∞).
Step 2
Write down the x-coordinate of the point of inflection of $g$.
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Answer
The x-coordinate of the point of inflection is at x = rac{2}{3}, where the second derivative changes sign.
Step 3
For which values of $x$ will $g$ be concave down?
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Answer
A function is concave down when its second derivative is negative. Given the nature of the cubic function and its turning points, g will be concave down for xextin(−1,2).
Step 4
If $g' (x) = -6x^2 + 6x + 12$, determine the equation of $g$.
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Answer
To find the equation of g, we integrate the derivative:
g(x) = rac{-6}{3}x^3 + rac{6}{2}x^2 + 12x + C = -2x^3 + 3x^2 + 12x + C
Given the y-intercept is 0, we find C=0. Thus, the equation of g is:
g(x)=−2x3+3x2+12x.
Step 5
Determine the equation of the tangent to $g$ that has the maximum gradient.
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Answer
To find the maximum gradient, we analyze the first derivative:
g′(x)=−6x2+6x+12
Setting this to 0 to find critical points,
−6x2+6x+12=0
Solving yields x = rac{1}{2} (maximum gradient). Substituting this back into the derivative gives:
m = g'( rac{1}{2}) = 27/2
Now, find the y value:
g( rac{1}{2}) = -2( rac{1}{2})^3 + 3( rac{1}{2})^2 + 12( rac{1}{2}) = 13.5 - 0.25 = 13.25
Thus, the tangent line's equation is:
y = rac{27}{2}x + c
To find c, use the point ( rac{1}{2}, 13.25) leading to the tangent equation being:
y = rac{27}{2}x - 3.25.
