Question

the graph shows the linear function y = 7x and the exponential function y = 7^x. which statement is true? the growth rate of the exponential function exceeds the growth rate of the linear function for -1 ≤ x ≤ 3. the growth rate of the exponential function exceeds the growth rate of the linear function for -1 ≤ x ≤ 1. the growth rate of the exponential function exceeds the growth rate of the linear function for 1 ≤ x ≤ 3. the growth rate of the exponential function exceeds the growth rate of the linear function for 0 ≤ x ≤ 3.

Explanation:

Step1: Analyze the functions

The linear function is \( y = 7x \), its slope (growth rate) is \( 7 \). The exponential function is \( y = 7^x \), its growth rate is the derivative (or we can analyze the rate of change by looking at the graph and function values).

Step2: Evaluate intervals

• For \( - 1\leq x\leq1 \):
• At \( x=-1 \), linear: \( y = 7\times(-1)=-7 \), exponential: \( y = 7^{-1}=\frac{1}{7}\approx0.14 \).
• At \( x = 0 \), linear: \( y = 0 \), exponential: \( y = 1 \).
• At \( x = 1 \), both are \( y = 7 \).
• The exponential function's rate of change (slope of tangent) at \( x = 0 \) is \( \ln(7)\times7^{0}=\ln(7)\approx2.01 \), which is less than \( 7 \) (slope of linear). But wait, maybe we should check the rate of change over the interval. The average rate of change of linear over \( [-1,1] \) is \( \frac{7 - (-7)}{1 - (-1)}=\frac{14}{2} = 7 \). The average rate of change of exponential over \( [-1,1] \) is \( \frac{7 - \frac{1}{7}}{1 - (-1)}=\frac{\frac{48}{7}}{2}=\frac{24}{7}\approx3.43 \), which is less than \( 7 \). So the linear growth rate exceeds exponential here? Wait, no, maybe I mixed up. Wait the question is "the growth rate of the exponential function exceeds the growth rate of the linear function" or vice versa? Wait no, let's re - read. Wait the options are "The growth rate of the exponential function exceeds the growth rate of the linear function" for different intervals. Wait no, wait the linear function is \( y = 7x \), slope \( 7 \). The exponential function \( y = 7^x \), its derivative is \( y'=7^x\ln(7) \).

• For \( 1\leq x\leq3 \):
• At \( x = 1 \), derivative of exponential is \( 7\ln(7)\approx7\times2.01 = 14.07 \).
• At \( x = 3 \), derivative of exponential is \( 7^{3}\ln(7)=343\times2.01\approx689.43 \).
• The slope of the linear function is always \( 7 \). So for \( x\geq1 \), the derivative of the exponential function (instantaneous growth rate) is greater than \( 7 \) (since at \( x = 1 \), \( 7\ln(7)\approx14.07>7 \) and it's increasing as \( x \) increases). So the exponential growth rate exceeds the linear growth rate for \( 1\leq x\leq3 \).

Answer:

The growth rate of the exponential function exceeds the growth rate of the linear function for \( 1\leq x\leq3 \) (the option corresponding to this statement).