Answer to Question #97980 in Calculus for Paul

Question

Find the roots of the function f(x) = 3^x · (log base2(x) − 3)^5 · e^x²−3x
.

Answer 1

$$f(x)=3^x(\log_2(x)-3)^5e^{x^2}-3x$$

Domain

$D(f): (0, \infin)$

If $0<x<8,$ then $\log_2(x)-3<0.$ Hence $f(x)<0$ for $0<x<8.$

Find the first derivative

$$f'(x)=3^x\cdot\ln3\cdot(\log_2(x)-3)^5e^{x^2}+$$ $$+5\cdot3^x\cdot{1 \over x\ln2}(\log_2(x)-3)^4e^{x^2}+$$ $$+2x\cdot3^x(\log_2(x)-3)^5e^{x^2}-3$$

Find the second derivative

$$f''(x)=3^x\cdot(\ln3)^2\cdot(\log_2(x)-3)^5e^{x^2}+$$ $$+5\cdot3^x\cdot{\ln3 \over x\ln2}\cdot(\log_2(x)-3)^4e^{x^2}+$$ $$+2x\ln3\cdot3^x(\log_2(x)-3)^5e^{x^2}+$$ $$+5\cdot3^x\cdot{\ln3 \over x\ln2}(\log_2(x)-3)^4e^{x^2}-$$ $$-5\cdot3^x\cdot{1 \over x^2\ln2}(\log_2(x)-3)^4e^{x^2}+$$ $$+20\cdot3^x\cdot{\ln3 \over x^2(\ln2)^2}\cdot(\log_2(x)-3)^3e^{x^2}+$$ $$+10\cdot3^x\cdot{1 \over \ln2}(\log_2(x)-3)^4e^{x^2}+$$ $$+2\cdot3^x(\log_2(x)-3)^5e^{x^2}$$ $$+2x\cdot\ln3\cdot3^x(\log_2(x)-3)^5e^{x^2}$$ $$+10\cdot3^x\cdot{1 \over \ln2}(\log_2(x)-3)^4e^{x^2}+$$ $$+4x^2\cdot3^x(\log_2(x)-3)^5e^{x^2}$$

If $x>8,$ then $f''(x)>0.$ Hence $f'(x)$ increases for $x>8.$

$f(8)=-24<0,$

$f'(8)=-3<0,$

$f''(8)=0.$

$f(8.0000000936436)\approx-24.0000$

$f'(8.0000000936436)=0.000011>0$

$f(8.00000498410413474204)\approx-0.0000000729$

$f(8.00000498410413474205)\approx0.00000004565$

The only root

$$x\approx8.00000498410413474205$$