Question

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Explanation:

Step1: Find critical points

Compute first derivative:
$f'(x) = 3x^2 - 6x - 9$
Set $f'(x)=0$:
$3x^2 - 6x - 9 = 0$
Divide by 3: $x^2 - 2x - 3 = 0$
Factor: $(x-3)(x+1)=0$
Solutions: $x=3, x=-1$

Step2: Find max/min y-values

For $x=-1$:
$f(-1) = (-1)^3 - 3(-1)^2 -9(-1)+10 = -1 -3 +9 +10 = 15$
For $x=3$:
$f(3) = 3^3 - 3(3)^2 -9(3)+10 = 27 -27 -27 +10 = -17$
Points: $(-1,15)$ and $(3,-17)$

Step3: Calculate line slope

Slope $m = \frac{-17-15}{3-(-1)} = \frac{-32}{4} = -8$

Step4: Find line equation

Use point-slope form with $(-1,15)$:
$y - 15 = -8(x + 1)$
Simplify: $y = -8x -8 +15$
$y = -8x +7$

Step5: Find inflection point

Compute second derivative:
$f''(x) = 6x -6$
Set $f''(x)=0$:
$6x-6=0 \implies x=1$
Find $y$-value:
$f(1) = 1^3 -3(1)^2 -9(1)+10 = 1-3-9+10 = -1$
Inflection point: $(1,-1)$

Step6: Verify point on line

Substitute $x=1$ into line equation:
$y = -8(1)+7 = -1$
This matches the inflection point's $y$-value.

Answer:

(a) The equation of the straight line is $\boldsymbol{y = -8x + 7}$
(b) The inflection point $(1,-1)$ satisfies the line equation, so it lies on the straight line.