Question

1. using the intermediate value theorem, show that the graphs of ( y = 5x^2 ) and ( y = e^x ) intersect for some ( x ) in (0,1).

Explanation:

Step1: Define the function

Let \( f(x)=5x^{2}-e^{x} \). We want to show that there exists a \( c\in[0,1] \) such that \( f(c) = 0 \) (since at the intersection point \( 5x^{2}=e^{x}\) implies \( 5x^{2}-e^{x}=0\)).

Step2: Check continuity of \( f(x) \)

The function \( y = 5x^{2}\) is a polynomial (hence continuous everywhere) and \( y=e^{x}\) is an exponential function (also continuous everywhere). The difference of two continuous functions is continuous, so \( f(x)=5x^{2}-e^{x}\) is continuous on the interval \([0,1]\).

Step3: Evaluate \( f(x) \) at the endpoints

• Evaluate \( f(0) \):

Substitute \( x = 0 \) into \( f(x) \): \( f(0)=5(0)^{2}-e^{0}=0 - 1=- 1\)

• Evaluate \( f(1) \):

Substitute \( x = 1 \) into \( f(x) \): \( f(1)=5(1)^{2}-e^{1}=5 - e\approx5 - 2.718 = 2.282\)

Step4: Apply Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( y = f(x) \) is continuous on a closed interval \([a,b]\) and \( k \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in the interval \([a,b]\) such that \( f(c)=k \).

Here, \( f(0)=- 1\) and \( f(1)\approx2.282\). We know that \( 0 \) is between \( f(0)=-1\) and \( f(1)\approx2.282\) (since \(-1<0<2.282\)). Since \( f(x) \) is continuous on \([0,1]\), by the Intermediate Value Theorem, there exists some \( c\in[0,1] \) such that \( f(c)=0 \), which means \( 5c^{2}-e^{c}=0\) or \( 5c^{2}=e^{c}\). So the graphs of \( y = 5x^{2}\) and \( y = e^{x}\) intersect for some \( x\in[0,1] \).

Answer:

By defining \( f(x)=5x^{2}-e^{x} \), showing \( f(x) \) is continuous on \([0,1]\), computing \( f(0)=-1\) and \( f(1)=5 - e\approx2.282\), and applying the Intermediate Value Theorem (since \( 0 \) is between \( f(0) \) and \( f(1) \)), we conclude there exists \( x\in[0,1] \) where the graphs intersect.