Determine how many critical values does have the function h(x)=definite integral of f(x) if f(x)=e^x*x^4 x>=0.limits of integration x=0 and x=x

We'll recall that the critical points of a function are
the zeroes of the first derivative of the
function.

Therefore, to determine the critical points of
h(x), we'll have to find out the first derivative of
h(x).

But h(x) = Int f(x)dx => h'(x) =
f(x)

Since h(x) is the definite integral of the function
f(x), we'll apply Leibniz Newton formula, to determine
h'(x).

h'(x) = f(x) - f(0)

But
h'(x) = 0 => f(x) - f(0) = 0 => f(x) =
f(0)

We'll calculate f(0) = (e^0)*(0) =
0

f(x) = 0

(e^x)*(x^4) =
0

Since e^x cannot be zero for any value of x => x^4
= 0

x = 0

Since
the derivative of the function h(x) has only one root, therefore the function has only
one critical value, x = 0.