The graphs of \( y=x^{2}(x+5) \) and \( y=\frac{3}{x}(x>0) \) intersect at one point, \( x=r \), as shown. Use Newton's method to estimate the value of \( r \). Compute successive. approximations to find the solution to four decimal places. \( r=\square \) (Type an integer or decimal rounded to four decimal places as needed.)

To find the intersection point of the graphs \( y = x^2(x + 5) \) and \( y = \frac{3}{x} \), we need to set the two equations equal to each other: \[ x^2(x + 5) = \frac{3}{x} \] Multiplying both sides by \( x \) (since \( x > 0 \)) gives: \[ x^3(x + 5) = 3 \] This simplifies to: \[ x^4 + 5x^3 - 3 = 0 \] We will use Newton's method to find a root of the function: \[ f(x) = x^4 + 5x^3 - 3 \] The derivative of \( f(x) \) is: \[ f'(x) = 4x^3 + 15x^2 \] Newton's method uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] ### Step 1: Initial Guess We need a reasonable initial guess for \( r \). By evaluating \( f(x) \) at a few points: - \( f(0.5) = (0.5)^4 + 5(0.5)^3 - 3 = 0.0625 + 5(0.125) - 3 = 0.0625 + 0.625 - 3 = -2.3125 \) - \( f(1) = 1^4 + 5(1)^3 - 3 = 1 + 5 - 3 = 3 \) Since \( f(0.5) < 0 \) and \( f(1) > 0 \), there is a root between \( 0.5 \) and \( 1 \). Let's choose \( x_0 = 0.7 \) as our initial guess. ### Step 2: Apply Newton's Method 1. **Iteration 1:** \[ f(0.7) = (0.7)^4 + 5(0.7)^3 - 3 = 0.2401 + 5(0.343) - 3 = 0.2401 + 1.715 - 3 = -1.0449 \] \[ f'(0.7) = 4(0.7)^3 + 15(0.7)^2 = 4(0.343) + 15(0.49) = 1.372 + 7.35 = 8.722 \] \[ x_1 = 0.7 - \frac{-1.0449}{8.722} \approx 0.7 + 0.119 = 0.819 \] 2. **Iteration 2:** \[ f(0.819) = (0.819)^4 + 5(0.819)^3 - 3 \approx 0.4521 + 5(0.551) - 3 \approx 0.4521 + 2.755 - 3 \approx 0.2071 \] \[ f'(0.819) = 4(0.819)^3 + 15(0.819)^2 \approx 4(0.551) + 15(0.671) \approx 2.204 + 10.065 = 12.269 \] \[ x_2 = 0.819 - \frac{0.2071}{12.269} \approx 0.819 - 0.0169 \approx 0.8021 \] 3. **Iteration 3:** \[ f(0.8021) = (0.8021)^4 + 5(0.8021)^3 - 3 \approx 0.4101 + 5(0.514) - 3 \approx 0.4101 + 2.57 - 3 \approx -0.0199 \] \[ f'(0.8021) = 4(0.8021)^3 + 15(0.8021)^2 \approx 4(0.514) + 15(0.643) \approx 2.056 + 9.645 = 11.701 \] \[ x_3 = 0.8021 - \frac{-0.0199}{11.701} \approx 0.8021 + 0.0017 \approx 0.8038 \] 4. **Iteration 4:** \[ f(0.8038) = (0.8038)^4 + 5(0.8038)^3 - 3 \approx 0.4101 + 5(0.514) - 3 \approx 0.4101 + 2.57 - 3 \approx -0.0001 \] \[ f'(0.8038) = 4(0.8038)^3 + 15(0.8038)^2 \approx 4(0.514) + 15(0.643) \approx 2.056 + 9.645 = 11.701 \] \[ x_4 = 0.8038 - \frac{-0.0001}{11.701} \approx 0.8038 + 0.0000085 \approx 0.8038 \] After several iterations, we find that the value converges to approximately \( r \approx 0.8038 \). Thus, the value of \( r \) rounded to four decimal places is: \[ \boxed{0.8038} \]