Find the point in the interval [5,9] at which the function $f(x)=17e^{3x}$ equals its average value on that interval.
So I've got my function as follows $f(x)=\frac{1}{4}(\frac{17}{3}(e^{27}-e^{15})$ So I've simplify it to $\frac{17e^{12}}{12}$
Now I believe I have to solve for $c$ so I've replace $x$ in my original function with $c$ so it looks like $17e^{3c}=\frac{17e^{12}}{12}$
So how would I go on from here, or am I doing the whole thing wrong.
Please Help!!
The average value is indeed $\frac{1}{12}(17)(e^{27}-e^{15})$. It does not simplify in a very useful way, though you may want to write it as $$\frac{1}{12}(e^{15})(e^{12}-1).$$ Now you want to solve the equation $$17e^{3c}=\frac{1}{12}(17)(e^{15})(e^{12}-1).$$ We might as well cancel the $17$'s. After that, take the logarithm of both sides. The logarithm can be written as $15+\ln(e^{12}-1)-\ln(12)$. Divide by $3$ to get $c$.