Question

Determine the value of k if the graph of f is shifted so that the values of x for which the new graph h(x）=（x+k)^3 - 4(x + k)^2 - 11(x + k）+ 30 decreases,
is in the interval -3 < x < 1 and 2/3

Answer 1

Answer:

k = 2

Step-by-step explanation:

The equation of function f(x) is:

$f(x)=x^3-4x^2-11x+30$

To determine the value of k if the graph of f(x) is shifted so that the values of x for which the new graph h(x) = (x+k)³ - 4(x+k)² - 11(x+k) + 30 decreases on the interval -3 < x < 1²/₃, we first need to find the interval on which f(x) decreases.

A function is decreasing when its first derivative is negative.

Differentiate f(x):

$f'(x)=3x^2-8x-11$

Now set f'(x) < 0:

$f'(x) < 0\\\\3x^2-8x-11 < 0\\\\3x^2+3x-11x-11 < 0\\\\3x(x+1)-11(x+1) < 0\\\\(3x-11)(x+1) < 0$

As f'(x) is a quadratic function with a positive leading coefficient, its graph will be an upward-opening parabola. Therefore, it will be negative (below the x-axis) between its x-intercepts. The x-intercepts of f'(x) are the x-values when f'(x) = 0. As f'(x) = (3x - 11)(x + 1), then:

$3x-11=0 \implies x=\dfrac{11}{3}\\\\x+1=0 \implies x=-1$

So, f(x) is decreasing on the interval:

$-1 < x < \dfrac{11}{3}$

The equation of the new function h(x) is:

$h(x)=(x+k)^3-4(x+k)^2-11(x+k)+30\\\\h(x)=f(x+k)$

When we add (or subtract) k units to the x-value of a function, it shifts the function k units to the left (or right). Therefore, function h(x) is a horizontal translation of function f(x).

Given that h(x) decreases on the interval -3 < x < 1²/₃, we can find the value of k by comparing this interval to where f(x) is decreasing:

$-1 < x < \dfrac{11}{3} \;\;\longrightarrow \;\;-3 < x < 1\frac23$

On comparing the two intervals, we can see that h(x) is f(x) shifted 2 units to the left, as evidenced by the left endpoint of the interval on which f(x) is decreasing, x = -1, becoming x = -3. When we shift a function k units to the left, we add k units to its x-value, so the value of k is:

$\Large\boxed{\boxed{k=2}}$