Help with sketching this graph of a function with these properties.

Question

f(4)=-2; f’(x)<0 for x<4, f’(4)<0 and f’(x)<0 for x>4

I figured out a point on the graph is (4,-2)

But I don’t know what the graph looks like, please help.

Thank you

Answer 1

import matplotlib.pyplot as plt

import numpy as np

# Create x values ranging from 0 to 8

x = np.linspace(0, 8, 100)

# Define the function f(x) = -x + 2

# This satisfies f(4) = -2 and has a constant negative slope

y = -x + 2

# Create the plot

plt.figure(figsize=(8, 6))

plt.plot(x, y, label="f(x)", color='blue', linewidth=2)

# Plot the specific point (4, -2)

plt.scatter(4, -2, color='red', s=100, zorder=5, label="f(4) = -2")

# Add annotations to explain the conditions

plt.annotate('f\'(x) < 0 (Decreasing)', xy=(2, 0), xytext=(1, 2),

arrowprops=dict(facecolor='black', shrink=0.05))

plt.annotate('f\'(x) < 0 (Decreasing)', xy=(6, -4), xytext=(5.5, -1),

arrowprops=dict(facecolor='black', shrink=0.05))

plt.text(4.2, -1.8, "(4, -2)", fontsize=12, color='red')

# Set up the chart layout

plt.axhline(0, color='black', linewidth=1) # x-axis

plt.axvline(0, color='black', linewidth=1) # y-axis

plt.grid(True, linestyle='--', alpha=0.7)

plt.title("Graph satisfying: f(4)=-2 and f'(x) < 0 everywhere")

plt.xlabel("x")

plt.ylabel("f(x)")

plt.legend()

# Show the plot

plt.show()