A curve with equation of the form `y=a x^4+b x^3+c x+d` has zero gradient at the point `(0,1)` and also touches the `x-` axis at the point `(-1,0)` then the value of `x` for which the curve has a negative gradient are: `xgeq-1` b. `x<1` c. `x<-1` d. `-1lt=xlt=1`

Options

A. `x gt-1`
B. `x gt1`
C. `x lt-1`
D. `-1 le x le1`

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the required values.

### Step 1: Establish the equation of the curve
The curve is given by the equation:
\[ y = ax^4 + bx^3 + cx + d \]

### Step 2: Use the condition that the curve passes through the point (0, 1)
Since the curve passes through the point (0, 1), we substitute \( x = 0 \) and \( y = 1 \) into the equation:
\[ 1 = a(0)^4 + b(0)^3 + c(0) + d \]
This simplifies to:
\[ d = 1 \]

### Step 3: Differentiate the equation to find the gradient
The derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 4ax^3 + 3bx^2 + c \]

### Step 4: Use the condition that the gradient is zero at (0, 1)
At the point (0, 1), the gradient is zero:
\[ 0 = 4a(0)^3 + 3b(0)^2 + c \]
This simplifies to:
\[ c = 0 \]

### Step 5: Use the condition that the curve touches the x-axis at (-1, 0)
Since the curve touches the x-axis at (-1, 0), we substitute \( x = -1 \) and \( y = 0 \):
\[ 0 = a(-1)^4 + b(-1)^3 + c(-1) + d \]
Substituting \( c = 0 \) and \( d = 1 \):
\[ 0 = a - b + 1 \]
This gives us our first equation:
\[ a - b + 1 = 0 \quad \text{(1)} \]

### Step 6: Find the gradient at the point (-1, 0)
We also need the derivative at this point to be zero (since it touches the x-axis):
\[ 0 = 4a(-1)^3 + 3b(-1)^2 + c \]
Substituting \( c = 0 \):
\[ 0 = -4a + 3b \quad \text{(2)} \]

### Step 7: Solve the system of equations
We now have two equations:
1. \( a - b + 1 = 0 \)
2. \( -4a + 3b = 0 \)

From equation (1), we can express \( b \) in terms of \( a \):
\[ b = a + 1 \]

Substituting \( b \) into equation (2):
\[ -4a + 3(a + 1) = 0 \]
\[ -4a + 3a + 3 = 0 \]
\[ -a + 3 = 0 \]
\[ a = 3 \]

Now substituting \( a = 3 \) back into \( b = a + 1 \):
\[ b = 3 + 1 = 4 \]

### Step 8: Write the final equation of the curve
Now we have:
- \( a = 3 \)
- \( b = 4 \)
- \( c = 0 \)
- \( d = 1 \)

Thus, the equation of the curve is:
\[ y = 3x^4 + 4x^3 + 1 \]

### Step 9: Find the conditions for negative gradient
To find where the gradient is negative, we set:
\[ \frac{dy}{dx} < 0 \]
\[ 12x^3 + 12x^2 < 0 \]
Factoring gives:
\[ 12x^2(x + 1) < 0 \]

### Step 10: Analyze the inequality
The critical points are \( x = 0 \) and \( x = -1 \). We analyze the sign of \( 12x^2(x + 1) \):
- For \( x < -1 \): \( 12x^2 > 0 \) and \( x + 1 < 0 \) → negative
- For \( -1 < x < 0 \): \( 12x^2 > 0 \) and \( x + 1 > 0 \) → positive
- For \( x > 0 \): \( 12x^2 > 0 \) and \( x + 1 > 0 \) → positive

Thus, the curve has a negative gradient for:
\[ x < -1 \]

### Final Answer
The value of \( x \) for which the curve has a negative gradient is:
**Option C: \( x < -1 \)**