Question 1[Maximum number: 3]Find h(x).Show Answerrecognition that h(x)=∫h′(x)h(x)=\int h^{\prime}(x)h(x)=∫h′(x)h(x)=14e4x+6x+c(h(1.5)=)14e4(1.5)+6(1.5)+c=105c=−4.85719…h(x)=14e4x+6x−4.86(=14e4x+6x+96−14e6)\begin{aligned} & h(x)=\frac{1}{4} \mathrm{e}^{4 x}+6 x+c \\ & (h(1.5)=) \frac{1}{4} \mathrm{e}^{4(1.5)}+6(1.5)+c=105 \\ & c=-4.85719 \ldots \\ & h(x)=\frac{1}{4} \mathrm{e}^{4 x}+6 x-4.86\left(=\frac{1}{4} \mathrm{e}^{4 x}+6 x+96-\frac{1}{4} \mathrm{e}^{6}\right) \end{aligned}h(x)=41e4x+6x+c(h(1.5)=)41e4(1.5)+6(1.5)+c=105c=−4.85719…h(x)=41e4x+6x−4.86(=41e4x+6x+96−41e6)Add to Test