The equation of a curve y=f(x)y = \mathrm{f}(x) is in the form of a cubic polynomial. The curve has a maximum point at (−1,0)(-1, 0) and passes through the point with coordinates (4,25)(4, 25). The tangent to the curve at x=3x = 3 is parallel to the line with equation y=24x+5y = 24x + 5. Find f(x)\mathrm{f}(x).

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Worked solution

6

1. Set up the general equation for a cubic polynomial and find its derivative.

Let the cubic polynomial be f(x)=ax3+bx2+cx+d\mathrm{f}(x) = ax^3 + bx^2 + cx + d.
The derivative is:
f′(x)=3ax2+2bx+c\mathrm{f}'(x) = 3ax^2 + 2bx + c

2. Use the coordinates of the maximum point to form two equations.

Since the curve has a maximum point at (−1,0)(-1, 0), it passes through (−1,0)(-1, 0) and its gradient there is 00:
f(−1)=0  ⟹  a(−1)3+b(−1)2+c(−1)+d=0\mathrm{f}(-1) = 0 \implies a(-1)^3 + b(-1)^2 + c(-1) + d = 0
−a+b−c+d=0— (1)-a + b - c + d = 0 \quad \text{--- (1)}

f′(−1)=0  ⟹  3a(−1)2+2b(−1)+c=0\mathrm{f}'(-1) = 0 \implies 3a(-1)^2 + 2b(-1) + c = 0
3a−2b+c=0— (2)3a - 2b + c = 0 \quad \text{--- (2)}

3. Use the other given point and the tangent gradient to form two more equations.

The curve passes through (4,25)(4, 25):
f(4)=25  ⟹  a(4)3+b(4)2+c(4)+d=25\mathrm{f}(4) = 25 \implies a(4)^3 + b(4)^2 + c(4) + d = 25
64a+16b+4c+d=25— (3)64a + 16b + 4c + d = 25 \quad \text{--- (3)}

The tangent at x=3x = 3 is parallel to y=24x+5y = 24x + 5, so its gradient is 2424:
f′(3)=24  ⟹  3a(3)2+2b(3)+c=24\mathrm{f}'(3) = 24 \implies 3a(3)^2 + 2b(3) + c = 24
27a+6b+c=24— (4)27a + 6b + c = 24 \quad \text{--- (4)}

4. Solve the system of linear equations to find $a, b, c$ and $d$.

Subtracting (2) from (4) to eliminate cc:
(27a+6b+c)−(3a−2b+c)=24−0(27a + 6b + c) - (3a - 2b + c) = 24 - 0
24a+8b=24  ⟹  3a+b=3  ⟹  b=3−3a— (5)24a + 8b = 24 \implies 3a + b = 3 \implies b = 3 - 3a \quad \text{--- (5)}

Substituting (5) into (2):
3a−2(3−3a)+c=0  ⟹  c=6−9a— (6)3a - 2(3 - 3a) + c = 0 \implies c = 6 - 9a \quad \text{--- (6)}

Subtracting (1) from (3) to eliminate dd:
(64a+16b+4c+d)−(−a+b−c+d)=25−0(64a + 16b + 4c + d) - (-a + b - c + d) = 25 - 0
65a+15b+5c=25  ⟹  13a+3b+c=5— (7)65a + 15b + 5c = 25 \implies 13a + 3b + c = 5 \quad \text{--- (7)}

Substituting (5) and (6) into (7):
13a+3(3−3a)+(6−9a)=513a + 3(3 - 3a) + (6 - 9a) = 5
13a+9−9a+6−9a=513a + 9 - 9a + 6 - 9a = 5
−5a+15=5  ⟹  −5a=−10  ⟹  a=2-5a + 15 = 5 \implies -5a = -10 \implies a = 2

Now find b,c,b, c, and dd:
b=3−3(2)=−3b = 3 - 3(2) = -3
c=6−9(2)=−12c = 6 - 9(2) = -12
From (1): −2+(−3)−(−12)+d=0  ⟹  7+d=0  ⟹  d=−7-2 + (-3) - (-12) + d = 0 \implies 7 + d = 0 \implies d = -7

5. State the final cubic polynomial.

Substitute the values of a,b,c,a, b, c, and dd back into the generic cubic polynomial:
f(x)=2x3−3x2−12x−7\mathrm{f}(x) = 2x^3 - 3x^2 - 12x - 7

Answer

f(x)=2x3−3x2−12x−7\mathrm{f}(x) = 2x^3 - 3x^2 - 12x - 7

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