Question

homework - semester exam study guide
question 1
consider the graph of the function $f(x) = -x^3 + 3x^2 + 1$
a) label the local minimum as a on the graph below.
b) label the local maximum as b on the graph below.
c) write down the interval where $f(x) > 0$
$-3x^2 + 6x$
d) sketch the tangent to the curve at $x = -1$ on the graph below.
e) write down the equation of the tangent at $x = -1$.

Explanation:

Part (c)

Step1: Find the derivative

The function is \( f(x) = -x^3 + 3x^2 + 1 \). The derivative \( f'(x) \) is found using the power rule. For \( -x^3 \), the derivative is \( -3x^2 \), for \( 3x^2 \) it is \( 6x \), and the derivative of the constant 1 is 0. So \( f'(x) = -3x^2 + 6x \).

Step2: Solve \( f'(x) > 0 \)

We need to solve the inequality \( -3x^2 + 6x > 0 \). First, factor out \( -3x \) (or factor the quadratic):
\( -3x^2 + 6x = -3x(x - 2) \) (wait, actually, factoring correctly: \( -3x^2 + 6x = -3x(x - 2) \)? Wait, no: \( -3x^2 + 6x = -3x(x - 2) \)? Let's check: \( -3x(x - 2) = -3x^2 + 6x \), yes. But let's factor out a -3: \( -3(x^2 - 2x) = -3x(x - 2) \). Now, to solve \( -3x(x - 2) > 0 \). Multiply both sides by -1 (remember to reverse the inequality sign): \( 3x(x - 2) < 0 \). Now, find the critical points: \( x = 0 \) and \( x = 2 \). These divide the number line into intervals: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \). Test a value in each interval:

• For \( (-\infty, 0) \), let's pick \( x = -1 \): \( 3(-1)(-1 - 2) = 3(-1)(-3) = 9 > 0 \), which does not satisfy \( 3x(x - 2) < 0 \).
• For \( (0, 2) \), let's pick \( x = 1 \): \( 3(1)(1 - 2) = 3(1)(-1) = -3 < 0 \), which satisfies the inequality.
• For \( (2, \infty) \), let's pick \( x = 3 \): \( 3(3)(3 - 2) = 9 > 0 \), which does not satisfy.

But wait, our original inequality was \( -3x(x - 2) > 0 \). So when we multiplied by -1, we reversed the inequality. So the solution to \( -3x(x - 2) > 0 \) is the interval where \( 3x(x - 2) < 0 \), which is \( (0, 2) \)? Wait, no, let's do it again. Let's solve \( -3x^2 + 6x > 0 \). Factor: \( -3x(x - 2) > 0 \). The coefficient of \( x^2 \) is negative, so the parabola opens downward. The roots are at \( x = 0 \) and \( x = 2 \). For a downward opening parabola, the inequality \( -3x(x - 2) > 0 \) (which is \( f'(x) > 0 \)) holds between the roots. So between \( x = 0 \) and \( x = 2 \), the derivative is positive. So the interval is \( (0, 2) \).

Part (e)

Step1: Find the point on the curve at \( x = -1 \)

First, find \( f(-1) \). Substitute \( x = -1 \) into \( f(x) = -x^3 + 3x^2 + 1 \):
\( f(-1) = -(-1)^3 + 3(-1)^2 + 1 = -(-1) + 3(1) + 1 = 1 + 3 + 1 = 5 \). So the point is \( (-1, 5) \).

Step2: Find the slope of the tangent at \( x = -1 \)

The derivative \( f'(x) = -3x^2 + 6x \). Substitute \( x = -1 \) into \( f'(x) \):
\( f'(-1) = -3(-1)^2 + 6(-1) = -3(1) - 6 = -3 - 6 = -9 \). So the slope \( m = -9 \).

Step3: Use point - slope form to find the equation of the tangent line

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-1, 5) \) and \( m = -9 \).
Substitute these values: \( y - 5 = -9(x - (-1)) \), which simplifies to \( y - 5 = -9(x + 1) \).
Expand the right - hand side: \( y - 5 = -9x - 9 \).
Add 5 to both sides: \( y = -9x - 9 + 5 \), so \( y = -9x - 4 \).

Part (c) Answer:

The interval where \( f'(x)>0 \) is \( (0, 2) \)

Part (e) Answer:

The equation of the tangent at \( x=-1 \) is \( y=-9x - 4 \)

Answer:

Step1: Find the derivative

The function is \( f(x) = -x^3 + 3x^2 + 1 \). The derivative \( f'(x) \) is found using the power rule. For \( -x^3 \), the derivative is \( -3x^2 \), for \( 3x^2 \) it is \( 6x \), and the derivative of the constant 1 is 0. So \( f'(x) = -3x^2 + 6x \).

Step2: Solve \( f'(x) > 0 \)

We need to solve the inequality \( -3x^2 + 6x > 0 \). First, factor out \( -3x \) (or factor the quadratic):
\( -3x^2 + 6x = -3x(x - 2) \) (wait, actually, factoring correctly: \( -3x^2 + 6x = -3x(x - 2) \)? Wait, no: \( -3x^2 + 6x = -3x(x - 2) \)? Let's check: \( -3x(x - 2) = -3x^2 + 6x \), yes. But let's factor out a -3: \( -3(x^2 - 2x) = -3x(x - 2) \). Now, to solve \( -3x(x - 2) > 0 \). Multiply both sides by -1 (remember to reverse the inequality sign): \( 3x(x - 2) < 0 \). Now, find the critical points: \( x = 0 \) and \( x = 2 \). These divide the number line into intervals: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \). Test a value in each interval:

• For \( (-\infty, 0) \), let's pick \( x = -1 \): \( 3(-1)(-1 - 2) = 3(-1)(-3) = 9 > 0 \), which does not satisfy \( 3x(x - 2) < 0 \).
• For \( (0, 2) \), let's pick \( x = 1 \): \( 3(1)(1 - 2) = 3(1)(-1) = -3 < 0 \), which satisfies the inequality.
• For \( (2, \infty) \), let's pick \( x = 3 \): \( 3(3)(3 - 2) = 9 > 0 \), which does not satisfy.

But wait, our original inequality was \( -3x(x - 2) > 0 \). So when we multiplied by -1, we reversed the inequality. So the solution to \( -3x(x - 2) > 0 \) is the interval where \( 3x(x - 2) < 0 \), which is \( (0, 2) \)? Wait, no, let's do it again. Let's solve \( -3x^2 + 6x > 0 \). Factor: \( -3x(x - 2) > 0 \). The coefficient of \( x^2 \) is negative, so the parabola opens downward. The roots are at \( x = 0 \) and \( x = 2 \). For a downward opening parabola, the inequality \( -3x(x - 2) > 0 \) (which is \( f'(x) > 0 \)) holds between the roots. So between \( x = 0 \) and \( x = 2 \), the derivative is positive. So the interval is \( (0, 2) \).

Part (e)

Step1: Find the point on the curve at \( x = -1 \)

First, find \( f(-1) \). Substitute \( x = -1 \) into \( f(x) = -x^3 + 3x^2 + 1 \):
\( f(-1) = -(-1)^3 + 3(-1)^2 + 1 = -(-1) + 3(1) + 1 = 1 + 3 + 1 = 5 \). So the point is \( (-1, 5) \).

Step2: Find the slope of the tangent at \( x = -1 \)

The derivative \( f'(x) = -3x^2 + 6x \). Substitute \( x = -1 \) into \( f'(x) \):
\( f'(-1) = -3(-1)^2 + 6(-1) = -3(1) - 6 = -3 - 6 = -9 \). So the slope \( m = -9 \).

Step3: Use point - slope form to find the equation of the tangent line

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-1, 5) \) and \( m = -9 \).
Substitute these values: \( y - 5 = -9(x - (-1)) \), which simplifies to \( y - 5 = -9(x + 1) \).
Expand the right - hand side: \( y - 5 = -9x - 9 \).
Add 5 to both sides: \( y = -9x - 9 + 5 \), so \( y = -9x - 4 \).

Part (c) Answer:

The interval where \( f'(x)>0 \) is \( (0, 2) \)

Part (e) Answer:

The equation of the tangent at \( x=-1 \) is \( y=-9x - 4 \)